Q 


TO  THE 


K 

PROBLEMS 

IN    ®I* 


SCHOOL 
PHYSICS 


If 

1  J- 

I  MEMORIAL 
Henry  Senger 

KEY  TO  THE  PROBLEMS 


IN 


AVERTS    SCHOOL    PHYSICS 


NEW  YORK  AND   CHICAGO 

SHELDON    AND    COMPANY 


M6 


COPYRIGHT,  1896, 
BY  SHELDON  AND  COMPANY, 


INMEMORIAM 


TYPOGRAPHY  BY  J.  S.  CCSHING  &  Co.,  NORWOOD,  MASS. 


TO  THE  TEACHER. 

IF  you  find  any  error  in  any  of  these  solutions,  you  will 
confer  a  favor  on  the  publishers  and  the  author  by  carefully 
writing  out  the  correction  and  sending  it  to  Dr.  Elroy  M. 
A  very,  No.  657  Woodland  Hills  Avenue,  Cleveland,  Ohio. 


926494 


KEY 


TO    THE 


PROBLEMS  IN  AVEKY'S  SCHOOL  PHYSICS. 


CHAPTER   I. 

Page  2O. 

8.  See  Exercise  11,  page  95. 

9.  The  number  of  inches  in  a  meter  is  given  on  page  18 ; 
39.37  inches.     The  number  of   millimeters  in  an  inch  is  the 
reciprocal  of  the  value  of  a  millimeter  as  given  on  the  same 
page ;  1  -=-  0.03937  =  25.39. 

10.  Four  times  250  cu.  cm.  equals  1  liter. 

11.  See  §  21. 

1.0567  qts.  =  1,000  cu.  cm. 

1  qt.  =  1,000  cu.  cm.  -=-  1.0567  =  946.34  cu.  cm. 

Page  23. 

1.  A  liter  =  1  cu.  dm.  =  1,000  cu.  cm.     One  cu.  cm.  of  pure 
water  weighs   1    gram ;    1,000   cu.  cm.  of  water   weigh   1,000 
grams,  or  1  Kg. 

2.  1,000  g.  x  1.8  =  1,800  g. 

3.  1,250  cu.  cm.  of  water  weigh  l,250g. 
1,250  g.  x  0.8  =  1,000  g.  or  1  Kg. 

4.  Since  a  liter  of  water  weighs  1,000  g.,  250  g.  of  water  is 
i  of  a  liter  of  water. 

5 


6  KEY  TO   THE  PROBLEMS 

5.  1  cu.  dm.  =  1,000  cu.  cm.     1  cu.  dm.  of  water,  therefore, 
'weighs.  1,000  g.  .-  1  Kg. 

6.  1  liter  =  1,000  cu.  cm.  ;  1  dl.  =  100  cu.  cm.     1  dl.  of  water 
weighs  100  times  1  g.  =  100  g.  =  1  Hg. 

7.  18  cu.  in.  x  19  x  20  =  6,840  cu.  in. 
6,840  -r-  231  =  29.6+,  the  number  of  gals. 

8.  25  cu.  cm.  x  35  x  75  =  65,625  cu.  cm. 
65,625  -T-  1,000  =  65.625,  the  number  of  liters. 

NOTE.  —  Exercises  7  and  8  are  intended  to  illustrate  the  advantage  of 
the  decimal  scale  as  used  in  the  metric  measures.  It  is  easier  to  divide 
by  1,000  than  to  divide  by  231. 

Page  37. 

11.  Be  sure  that  the  pupil  understands  that  he  is  testing 
the  stiffness  and  not  the  strength  of  the  beams  used,  (a)  The 
deflection  is  proportional  to  the  load.  (&)  The  deflection  is 
proportional  to  the  cube  of  the  length,  (c)  The  deflection 
is  inversely  proportional  to  the  breadth,  (d)  The  deflection  is 
inversely  proportional  to  the  cube  of  the  thickness,  i.e.,  depth. 


13.  The  elongation  is  proportional  to  the  length  of  the  wire, 
and  inversely  proportional  to  the  cross-section  of  the  wire,  i.e., 
to  the  square  of  the  diameter. 


14.  The  angle  of  torsion  is  proportional  to  the  force  of  tor- 
sion. 

15.  The  angle  of  torsion  is  proportional  to  the  length  of  the 
rod. 

Page  42. 

4.  Air  held  in  solution  is  largely  expelled  by  boiling.  Fewer 
air  bubbles  will,  therefore,  accumulate  on  the  inner  wall  of  the 
tumbler  that  contains  the  boiled  water.  See  the  next  exercise. 


IN  AVERY'S  SCHOOL   PHYSICS.  7 

Page  54. 

6.  Porosity. 

7.  No.     Scientifically  speaking,  pores  are  the  minute  inter- 
stices between  molecules  ;  intermolecular  spaces. 

8.  Fluids  include  liquids.     Both  terms  imply  freedom  of 
molecular  motion,  but  the  tendency  of  the  molecules  to  cohere 
pertains  necessarily  to  liquids,  and  not  necessarily  to  fluids. 

9.  Water  and  steam  are  identical  substances,  varying  only 
in  respect  to  their  physical  condition  (§  41).     The  nature  of 
a  substance  is  determined  by  the  nature  of  its  molecules  (§  6) 
and  a  physical  change,  as  from  water  to  steam,  does  not  change 
the  molecule  (§  9).     As  there  is  no  chemical  change  in  passing 
from  water  to  steam,  no  molecule  is  changed  in  any  way ;  they 
remain  equal  in  size. 

10.  A  cubic  inch  of  water  makes  about  a  cubic  foot  of  steam. 
As  there  is  no  increase  in  the  size  or  the  number  of  the  mole- 
cules, the  increased  volume  must  be  due  to  an  increase  in  the 
distances  between  them. 

12.  See  §  45. 

13.  See  Experiment  36. 

14.  See  §  33  (a).     Extension,  impenetrability,  weight,  inde- 
structibility, inertia,  etc.,  pertain  to  all  matter  and  are,  there- 
fore, universal  properties.      Hardness,  malleability,  ductility, 
etc.,  pertain  to  only  certain  forms  of  matter  and  are,  therefore, 
characteristic  of  those  that  possess  them. 

15.  See  §  52  (6).     Dialysis  and  evaporation. 

16.  Divide  the  number  of  pounds  by  2.2046.     See"  §  25. 

18.  n  cc  — 

I 

19.  See  §  33  (a).    Tenacity  is  proportional  to  sectional  area, 
and  that  is  proportional  to  the  square  of  the  diameter. 

(14.19)2 :  (17.90)2 :  :  15  :  x. 


8  KEY  TO   THE  PROBLEMS 

See  Appendix  4.  The  squares  of  the  diameters  in  mils  may 
be  found  in  the  column  headed  circular  mils,  and  the  pro- 
portion written  directly  from  the  table,  thus  : 

201.5  :  320.4  : :  15  :  x. 

20.  From  the  table  of  wire  dimensions,  it  appears  that  No.  27 
wire  has  a  sectional  area  twice  that  of  No.  30  wire.  Conse- 
quently, No.  27  iron  wire  has  a  breaking  strength  of  10  pounds. 
Comparing  these  brass  and  iron  wires  of  the  same  size  (No.  27), 
it  appears  that  the  tenacity  of  spring-brass  is  1.5  times  that  of 
annealed  iron. 

Page  55. 

1.  Steam  passes  through  the  cardboard  and  condenses  upon 
the  inner  surface  of  the  upper  tumbler,  thus  illustrating  the 
porosity  of  the  cardboard. 


IN   AVERY'S   SCHOOL   PHYSICS.  9 


CHAPTER   II. 

Page  63. 

1.  500x500  =  250,000. 

2.  321.6  x  200  =  64,320. 

3.  |       X  9  ~  2()  |      Their  momenta  are  equal. 

4.  Uniformly  accelerated  motion,  like  that  of  a  falling  body 
which  is  acted  upon  by  the  attraction  of  the  earth. 

5.  There  are  5,280  feet  in  a  mile. 

5,280  x  15  x  1  _  g 

1,320  x  12 
The  momentum  of  the  ball  will  be  5  times  that  of  the  stone. 

(  50,000  x    2  =  100,000  ) 
,6'    {loioOOx  10  =  100,000}     Their  momenta  are  equal. 

7.  25  x  60  =  1,500,  the  momentum  of  the  first,  and,  conse- 
quently, of  the  second.     1,500  -*-  40  =  37.5.     The  velocity  of 
the  second  is  37.5  ft.  per  second. 

8.  100x20  =  2,000. 

2,000  -r-  500  =  4,  the  number  of  kilograms. 

9.  From  a  point,  (7,  draw  three  lines  3^  inches,  5  inches, 
and   6   inches    respectively   in   length.     Draw   the   first   line 
toward  the  right  hand;   the  second  line  downward;  and  the 
third  line  downward  and  toward  the  left  so  as  to  make  an 
angle    of    45°   with    the    second    line.       Eemember   the   old 
geographical  rule:  "The  top  of  the  map  is  north;  the  bottom, 
south;  the  right  hand,  east;  and  the  left  hand,  west."     The 
three  arrowheads  point  from  C. 


10  KEY  TO  THE  PROBLEMS 

10.  4^  x  5?ff°  =  58|,  the  velocity  in  feet  per  second. 
bO  X  t>0 

^    92,390,000  x  2 
(16  x  60)  +  36* 

12.  See  Formula  (3)  on  page  59. 

J=  Jo*8  =  10  ft.  x62  =  360ft., 
the  distance  traversed  in  6  seconds. 
See  Formula  (2)  on  page  58. 

v  =  at  =  20  ft.  x  6  =  120  ft.,  the  final  velocity. 

13.  The  load  of  50  Kg.  =  2.2046x50 -110.23  Ibs.  (see  §  25). 
This  load  is  7.35  times  as  great  as  the  load  that  can  be  carried 
by  the  No.  27  wire.     We  must,  therefore,  have  ajwire  that  is 
7.35  times  as  great  in  sectional  area.     From  the  table  of  wire 
dimensions,  we  find  that  the  diameter  of  No.  27  wire  is  0.01419 
of  an  inch.  1 :  7.35  :  :  (0.01419)2 :  ^ 

x  =  0.03847  of  an  inch,  the  diameter  required.  This  is 
greater  than  the  diameter  of  No.  19  wire  and  less  than  that  of 
No.  18  wire.  An  easier  solution  for  practical  purposes  is  to 
multiply  201.5,  the  number  of  circular  mils  for  No.  27  wire, 
by  7.35,  the  ratio  between  the  given  loads.  This  gives  1,481 
circular  mils,  which  is  greater  than  that  for  No.  19  wire  and 
less  than  that  for  No.  18  wire.  No.  18  wire  must  be  used. 

Page  64. 

1.  Draw  two  diameters  at  right  angles,  thus  dividing  the 
circumference  into  arcs  of  90°  each.  From  the  extremities  of 
these  diameters  as  centers,  draw  arcs  cutting  the  circumference 
into  arcs  of  30°  each.  Use  the  protractor  only  for  the  trisec- 
tion  of  the  30°  arcs. 

Page   68. 

1.  The  locomotive  must  travel  360  ft.  +  120  ft.  =  480  ft. 
It  moves  ^  mile  or  2,640  ft.  per  minute,  or  44  ft.  per  second. 
480  -r-  44  =  11  nearly,  the  number  of  seconds. 


IN  AVERY'S  SCHOOL  PHYSICS.  11 

2.  See  §  66  (a).     32.16  x  25  =  804,  the  number  of  poundals. 

3.  980  x  5,000  =  4,900,000,  the  number  of  dynes. 

4.  13,825   dynes.      Multiply   the    number   of    grams    in   a 
pound  by  the  number  of  centimeters' in  a  foot. 

5    64  x  16  x  1,300  __  1 109i     The  momentum  of  the  cannon- 

1  x  1,200 
ball  is  1,1091  times  that  of  the  bullet. 

6.  See  solution  to  Laboratory  Exercise  11,  page  37.     The 
deflection  is  proportional  to  the  cube  of  the  length. 

33 :  43 : :  0.5  :  x. 

7.  Inertia. 

Page    79. 

1.  Adopt  any  convenient  scale,  as  0.1   of  an  inch   to   the 
pound.     Then  the  100-pound  force  will  be  represented  by  a 
line  10  in.  long,  and  the  other  force  by  a  line  15  in.  long.     See 
§  69  (1).     The  resultant  will  be  represented  by  a  line  25  in. 
long. 

2.  See  §  69  (2).     Using  the  same  scale   as   in  Exercise  1, 
the  resultant  will  be  represented  by  a  line  5  in.  long,  and  the 
motion  will  be  in  the  direction  of  the  greater  force. 

3.  Suppose  we   adopt  the   scale   of  an   inch   to   the   mile. 
Draw  a  horizontal  line  4  inches  long  to  represent  the  force 
of  the  oars.     From  one  end  of  this  line,  draw  a  vertical  line 
3  inches  long  to  represent  the  force  of  the  current.     Join  the 
free- ends  of  these  lines;    the  hypotenuse  thus  formed  will 
represent  the  resultant  of  these  two  forces. 

32  +  42  =  25.     V25  =  5. 

See  §  70  (a).  The  boat  will  move  in  the  direction  indicated 
by  the  hypotenuse  and  with  a  velocity  of  5  miles  per  hour. 
Of  course,  the  problem  means  that  the  boat  is  headed  directly 
across  the  stream. 

4.  Draw  a  vertical  line,  64  units  (as  mm.,  or  16ths  of  an 
inch)  long.     From  the  foot  of  this  line  draw  a  horizontal  line 


12 


KEY   TO   THE   PROBLEMS 


24  units  long.  (Use  the  same  kind  of  units  that  you  adopted 
for  the  vertical  line.)  Join  the  free  ends  of  these  lines.  The 
hypotenuse  will  be  the  graphic  representation  of  the  re- 
sultant. 

642  +  242  =  4,672.     V4,672  =  68  + . 

5.  Draw,  as  before,  a  vertical  line  (3  x  20  = )  60  units  long, 
and  a  horizontal  line  (12  x  20  = )  240  units  long.     Draw  the 
hypotenuse. 

602  +  2402  =  61,200.     V6pOO  =  247  + . 

6.  Draw  BN=10  units  of  length.     From  B,  draw  BE  at 
right  angles  to  BN  and  make  it  15  units  long.     Complete  the 


parallelogram  and  draw  the  partial  resultant,  Br.  From  B, 
draw  BS  at  an  angle  of  45°  from  BE,  and  make  it  25  units 
long.  Complete  the  parallelogram,  and  draw  the  diagonal, 
BM,  which  will  represent  the  complete  resultant.  The  scale 
here  adopted  is  2  mm.  per  pound.  The  line,  BR,  being  67  mm. 
long,  represents  a  force  of  33|-  pounds.  The  direction  of  the 
resultant  is  a  little  to  the  north  of  west,  as  is  indicated  by 
the  angle,  EBR. 


IN  AVERY'S   SCHOOL  PHYSICS.  13 

7.  See  §  60  (3).     The  momentum  of  the  gun  is  equal  to 
the  momentum  of  the  projectile.     As  the  gun  is  heavier  than 
the  projectile,  its  velocity  must  be  less  than  that  of  the  pro- 
jectile, in  order  that  the  products  of  the  numbers  representing 
the  weight  and  velocity  in  each  case  may  be  equal. 

8.  The  width  of  the  river  is  represented  by  a  line  4  units 
long ;  the  actual  course  of  the  boat  by  a  line  5  units  long.     If 
the  4  units  represent  1  mile,  the  5  units  will  represent  1^  miles, 
the  distance  that  the  boat  moves.   It  takes  no  longer.    See  §  62. 

9.  See  Fig.  94,  in  which  LM  represents  the  plank,  and  MN 
the  distance  that  one  end  of  it  is  raised.      WC  represents  the 
weight  of  the  cask.     From  T7,  draw  WD  perpendicular  to  LM. 
From  (7,  draw  CD  parallel  to  LM.    Complete  the  parallelogram, 
WBCD.      The  force  of  gravity  represented  by    WC  may  be 
resolved  into  two  components,  represented  by  WD  and  WB. 
WB  represents  the  force  with  which  the  cask  tends  to  roll 
down  the  plank.     This  tendency  may  be  successfully  resisted 
by  a  force  represented  by  WB',  equal  to  WB  and  opposite  in 
direction.      WB  =  ^  WC,  as  may  be  seen  by  direct  measure- 
ment or  as  may  be  proved  geometrically,  the  triangles  WBC 
and  LNM  being  similar.     Hence,  the  muscular  force  needed  is 
25  Ibs.  or  more. 

10.  See    §   66  (a).      32.16  x  60  =  1,929.6,    the    number    of 
poundals. 

11.  60  Kg.  =  60,000  g. 

980  x  60,000  =  58,800,000,  the  number  of  dynes. 

12.  See  §  65,  and  §  66  (a). 

32.16  x  10  =  321.6,  the  number  of  poundals. 

13.  1,000  -f-  20  =  50,  the  number  of  grams. 

14.  12  x  6  =  72,  the  number  of  dynes. 

15.  490  -j-  70  =  7,  the  number  of  centimeters. 

16.  The  wind  strikes  obliquely  upon  the  faces  of  the  blades 
and  is  resolved  by  the  blades  into  two  components,  as  it  is  in 


14  KEY  TO  THE  PROBLEMS 

the  case  of  the  obliquely  set  sails  of  vessels.  One  of  these 
components  tends  to  move  the  blade  in  a  direction  perpendic- 
ular to  that  of  the  wind.  The  corresponding  blade  on  the 
opposite  side  of  the  wheel  is  similarly  urged  in  the  opposite 
direction,  so  that  the  force  of  the  wind  upon  each  such  pair  of 
blades  constitutes  a  couple  that  helps  to  turn  the  wheel  about 
its  axis. 

17.   See  §  78. 


r 

18.  Compute  the  length  of  the  hypotenuse  of  an  imaginary 
right-angled  triangle  with  sides  of  5  and  12  respectively.     See 
§  70  (a).  52  +  122  =  0.  169  =  a-2.  13  =  3.. 

The  resultant  has  a  force  of  13  Kg. 

19.  In  this  case,  the  hypotenuse  of  the  right-angled  triangle 
is  30,  and  one  of  the  other  sides  is  18. 

302  -  182  =  x\ 
x  —  24,  the  numerical  value  of  the  component  sought. 

20.  See  Experiment  45,  and  §  69  (3). 

8  +  11  =  19. 
19:    8::57:24. 
19  :  11  :  :  57  :  33. 

The  point  will  be  24  inches  from  the  line  of  the  11-pound  force, 
and  33  inches  from  the  line  of  the  24-pound  force. 

21.  See  §  58.     The  southward  and  the  eastward  motions 
are  equal.     The  parallelogram  is  a  square  with  a  diagonal  of 
30  units. 


V450  =  21+,  the  number  of  miles. 
22.   See  §  78. 

G.F.  =  ^10,000  x  (2,000)*  =  40,000,000,   the  nmnber   of 

r  1,000 

dynes. 


IN  AVERY'S  SCHOOL  PHYSICS.  15 

23.  (a)  See  §  60  (3).  The  action  on  the  sail  will  be  balanced 
by  the  reaction  on  the  bellows  and  the  boat  will  not  be  moved 
by  the  operation.  (6)  The  reaction  on  the  bellows  will  push 
it  and  the  boat  ahead. 

Page  81. 

1.  See  §  69  (3). 

2.  From  the  ends  of  a  vertical  line  5  inches  long,  draw 
toward  the  right  hand  horizontal  lines  3^-  and  5^  inches  long 
respectively.     An  arrowhead  on  each  line  should  point  to  the 
right.     From  a  point  on  a  vertical  line,  T7¥  of  its  length  from 
the  5|-inch  line  and  -j-J-  of   its  length  from  the  3 J -inch  line, 
draw  a  horizontal  line  to  the  left  and  9  inches  long.    This  line 
with  an  arrowhead  pointing  to  the  right  represents  the  result- 
ant.    The  same  line  with  an  arrowhead  pointing  to  the  left 
represents  the  equilibrant. 

3.  The  line  drawn  downward  from  0  with  a  length  equal 
to  OD  represents  the  gravity  of  the  unknown  weight.     01) 
represents  the  equilibrant  of  that  weight. 

4.  ZT  represents  the  equilibrant  of  the  10-pound  pull  of 
the  weight.     The  scale  adopted  is  -J-  inch  =  1  pound.    See  §  70. 

7.  Draw  an  equilateral  triangle  as  follows :  lay  off  a  line 
AB,  of  convenient  length.     From  A  as  the  center  and  with  a 
radius  equal  to  AB,  describe  an  arc.     From  B  as  a  center  and 
with  the  same  radius,  describe  a  second  arc  that  cuts  the  first 
arc.     Mark  the  point  of  intersection  of  the  two  arcs,  C,  and 
join  AC  and  BC. 

8.  The  angle  measures  127°  3'+. 

Page  9O. 

1.  See  §  81.      8'250  x  176  =  n  the  number  of  H.  P. 

550  x  60  x  4 

SUGGESTION  :  Reduce  by  cancellation. 

2.  gee  §85.    192'96  x  10'000  =  30,000. 

64.32 


16  KEY  TO' THE  PROBLEMS 

3.   The  direction  makes  no  difference. 
50  x  19.6  x  19.6 


=  50  x  19.6  =  980,  the  number  of    kilo- 
A  x  y.o 

gram-meters. 

4.  50  x  750  =  the  momenta. 

Kf)    v   7KA 

*         =  500,  the  required  velocity  in  feet. 

75 

5.  The  velocity  is  500  m.  per  minute,  or  8^  m.  per  second. 
40  x  81  x  81  =  141|^  the  number  of  kilogram-meters. 

£  X  t/.o 

6.  1?50°  X  2'876  =  36,  the  number  of  H.  P. 
550  x  60  x  3 

7.  That  quantity  of  water  weighs  about  (62.5  Ibs.  x  300  =) 
18,750  pounds. 

=  75,  the  number  of  H.  P.  necessary. 


oo 
00,000 

8.   A  velocity  of  20  miles  per  hour  is  one  of  29J  feet  per 
second. 


K.E.  =  1QO  X  29^  X  29^  =  the  number  of  foot-pounds. 
u4.o^ 

9.   (a)  6,000  x  50  =  300,000,  the  number  of  foot-pounds. 
(6)  300,000  --  16,500  =  the  number  of  H.  P. 

10.  2  =  10>QQQ  X  1QQ.    ...  m  =  15  5    the  number  of  minutes. 

33,000m 

11.  2=  10>000^.     a?  =  3.3,  the  number  of  feet. 

o50  x  30 

12.  It  is  often  convenient  to  remember  that  550  foot-pounds 
per  second  is  equivalent  to  33,000  foot-pounds  per  minute. 
1,650,000  -*-  33,000  =  50,  the  number  of  H.  P. 


IN   AVERY'S  SCHOOL  PHYSICS.  17 

14.  K.  E.  =  2°*  X0f  Q°  =  1,250,  the   number   of  foot-pounds 

o4.oJ 

that  the  moving  sphere  can  perform.  This  working  power  is 
the  exact  measure  of  the  work  performed  upon  it. 

15.  A  resistance  of  8  pounds  per  ton  signifies  that  to  move 
a  ton  one  foot  on  the  rails  involves  as  much  work  as  to  lift  8 
pounds  one  foot  high,  or  8  foot-pounds.     To  move  10  tons  50 
ft.  on  the  rails  would  require  4,000  foot-pounds.     The  addi- 
tional work  done  in  giving  kinetic  energy  to  the  car  will  be 
measured  by  the  kinetic  energy  of  the  car.     The  velocity  of 
the  car  is  4.4  ft.  per  second. 

KE-=  20,000^  44  x  4.4  =  6>019+> 

the  number  of  foot-pounds  that  the  car  can  perform,  or  the 
amount  of  work  done  in  giving  the  car  the  velocity  specified. 
4,000  +  6,019  =  10,019,  the  total  amount  of  work,  in  foot- 
pounds, that  was  done  on  the  car. 

16.  The  triangle  or  parallelogram  will  not  "close."     None 
of  the  given  forces  can  be  the  equilibrant  of  the  other  two. 

17.  See  §  46  (a). 

18.  The  velocity  along   the   track   is   29^   ft.    per   second. 
The   problem   is   to   find    the   hypotenuse    of    a   right-angled 
triangle,  the  sides  of  which  measure  respectively  29J  and  20 
ft.     This  magnitude  of  35.5  ft.  represents  the  velocity  per 
second. 

19.  15  x  20  -f-  4  =  75,  the  number  of  dynes. 

20.  The  momenta  of    gun   and   projectile  =  250  x  1,420  = 
355,000.     This  divided  by  34,000,  the  weight  of  the  gun,  gives 
14.79  as  the  required  velocity  in  feet. 

21.  6  x  3.14159  -r-  5  =  3.769,  the  velocity  in  feet  per  second. 

The  C.  F.  =  2Q  x  (3-769)2.     gee  §  78. 
3 


18  KEY  TO  THE  PROBLEMS 

22.  50  x  10  =  500,  the  number  of  foot-pounds. 

23.  The  force  is  resolved  into  two  components,  one  parallel 
to  the  track  and  producing  motion ;  the  other  perpendicular  to 
the  track  and  producing  no  motion.     The  first  component  is 
the  only  one  that  overcomes  resistance  or  does  work.    See  §  79. 
In   the  parallelogram   constructed  for   the   resolution  of   the 
force,  we  have  50  as  the  diagonal  of  a  square,  the  length  of 
one  side  of  which  is  to  be  determined. 

KA2 

—  =  35.355,  the  force  in  pounds, 

35.355  x  10  =  353.55,  the  work  in  poundals. 

24.  His  push  contributes  nothing  to  the  motion  of  the  car, 
and  so  it  does  no  work  on  the  car.     See  §  79. 

25.  The  man  does  not  push  the  car  or  help  to  push  it ;  the 
car  pushes  the  man.     The  man  does  no  work  upon  the  car; 
the  car  does  work  upon  the  man.     The  exercise  is  the  converse 
of  Exercise  23. 

26.  (a)  See  §  85.      50  x  30-  foot-pounds. 

32.16  x  2 

(6)  See  §  86.     50  *3°2  foot-poundals. 

1,056,000         =  1 
'  550  x  60  x  60  x  8     15* 

Page  92. 

1.  It  is  better  to  take  a  tenth  of  the  number  of  revolutions 
made  by  the  spindle  while  the  large  wheel  revolves  10  times. 
Eecord  this  ratio  on  the  whirling  table. 

2.  The  greater  the  speed,  the  greater  the  centrifugal  force 
of  the  ball,  and  the  greater  the  length  to  which  the  cord  is 
stretched. 

3.  The  mercury  gets  as  far  as  possible  from  the  center  of 
rotation,  and  thus  rises  to  the  upper  ends  of  the  tubes. 


IN  AVERY'S   SCHOOL  PHYSICS. 


19 


4.  In  the  geologic-long-ago,  when  the  earth  was  in  a  plastic 
condition,  its  rotation  upon  its  axis  would  give  it  the  spheroidal 
form  that  it  now  has ;  it  would  flatten  out  at  the  poles. 

7.  See  §  69  (2).     The  components  are  equal  and  the  result- 
ant is  zero. 

8.  This  exercise  shows  that  centrifugal  force  varies  directly 
as  the  mass  and  inversely  as  the  radius,  as  is  indicated  by  the 
formula  given  in  §  78. 

10.  The  momenta  produced  by  the  elastic  force  of  the  cord 
are  equal ;  the  lighter  block  will  move  twice  as  fast  as  the 
other  block. 

11. 


Page   98. 

2.  As  the  assumed  distance  from  the  earth's  center  is  only 
one-fourth  of  the  surface  distance,  the  required  weight  will  be 
only  one-fourth  of  the  surface  weight.  550  -*-  4  =  137^.  Or, 
4,000  : 1,000  : :  550  :  x. 


20  KEY  TO  THE  PROBLEMS 

NOTE.  —  The  teacher  will  be  fortunate  who  finds  that  all  of  his  pupils 
are  able  to  handle  a  proportion  intelligently  and  easily.  Secure  this 
ability  if  possible.  Frequently  vary  the  form  of  the  statement  from 

a  :  b  =  C  :  d,  to  2  =  1 


=  = 

502     2,500 

4.  If  the  first  and  second  were  at  equal  distances  from  the 
third,  the  first  would  have,  on  account  of  its  lesser  mass,  only 
-§-  or  -|  as  great  an  attraction  as  the  second.  But  being  only 
half  as  far  distant,  its  attraction  is  four  times  as  great  as  it 
would  be  if  it  were  at  an  equal  distance.  -f  x  4  =  -|.  The 
smaller  ball  exerts  2J  as  much  force  upon  the  third  ball  as 
the  larger  one  does. 

502     8  (9:6) 


5.  Beckon  distances  from  the  earth's  center. 

12,0002  :  4,0002  :  :  900  :  x. 

Or,  the  distance  from  the  earth's  center  being  increased  three- 
fold, the  weight  will  be  divided  by  32  or  9. 
900  Ibs.  -s-  9  =  100  Ibs. 

6.  1  :  16  :  :  4,0002  :  or8.     .-.  a  =  16,000,  the  number  of  miles 
from  the  earth's  center.     16,000  -  4,000  =  12,000,  the  number 
of  miles  from  the  earth's  surface. 

7.  7,0002  :  4,0002  :  :  200  :  x.     The  man  would  weigh  65.3  Ibs. 
7,0002  :  4,0002  :  :  100  :  y.     The  boy  would  weigh  32.65  Ibs. 

8.  4,0002  :  6,0002  :  :  180  :  x.     Ans.  80  Ibs. 

9.  Work  as  in  preceding  examples,  or  as  follows: 

50  Ibs.  x  if  =  32  Ibs.,  the  weight  1,000  miles  above  the 
surface. 

50  Ibs.  x  f  =  371  Ibs.,  the  weight  1,000  miles  below  the 
surface. 

It  would  weigh  5J  Ibs.  more  when  below  the  surface. 


IN  AVERY'S  SCHOOL   PHYSICS.  21 

10.  It  would  be  ^  as  great  in  either  case. 

11.  See  solution  to  Exercise  23,  page  92.      The   effective 
component  is  a  force  of  565.685  pounds. 

565.685  x  12  =  6,788.22,  the  number  of  foot-pounds. 

12.  1,200  :  2,700  : :  4,0002 :  x2.     .-.  x  =  6,000. 
6,000  -  4,000  =  2,000. 

13.  It  would  increase  the  weight  fourfold.     See  §  90. 

14.  In  such  cases  of  mutual  action  between  free  bodies,  the 
two  bodies  experience   equal  changes    of   momentum.      Just 
before  collision,  the  small  ball  had  a  momentum  of  3,750  in 
a  certain  direction.     Just  after  collision,  it  had  a  momentum 
of  1,250  in  the  opposite  direction.     One  might  be  called  posi- 
tive and  the  other  negative.     The  change  of  momentum  of  the 
small   ball  was   5,000.      This   change  was   produced   by  the 
reaction  of  the  collision.     The  equivalent  action  must  have 
given  the  large  ball  a  momentum  of  5,000.     This  divided  by 
the  weight  of  the  large  ball  gives  the  velocity  of  the  large  ball. 
5,000  -r-  200  =  25,  the  required  velocity  in  feet  per  second. 

Page    1O3. 

1.  See  §§  97,  100. 

2.  See  §  97. 

3.  See  §  100  (6). 

4.  The  center  of  mass  is  lower.     See  §  100. 

5.  The  distance  from  the  center  of  the  earth  being  multi- 
plied by  60,  the  attraction  will  be  divided  by  602.     See  §  90. 

32.16  ft.  -s-  3,600  =  0.0089  +  ft. 

6.  Draw  two  lines,  respectively  10  and  7.5  cm.  in  length, 
making  them  include  an  angle  of  45  degrees.     Complete  the 
parallelogram    and   draw  the   diagonal  from   an  acute  angle. 
Its  length  of  12.147  cm.  represents  a  value  of  121.47  feet  per 
second.     The  direction  is  25°  53'  to  the  east  of  north. 


22  KEY  TO   THE  PROBLEMS 

7.  5,280  ft.  -j-  60  =  88  ft.,  the  velocity  per  second.     This 
velocity  was  developed  in  600  seconds.      88  -r-  600  =  0.1466, 
the  acceleration  in  feet.    Force  =  mass  x  acceleration;  /—  ma. 
§§  66  (6)  and  86. 

0.1466  x  100  =  14.66,  the  number  of  poundals. 

8.  See  §  81. 


Or, 

1  H.  P.  =  550  foot-pounds  in  1  second. 
100  H.  P.  =  55,000  foot-pounds  in  1  second. 

=  3,300,000  foot-pounds  in  1  minute. 
=  198,000,000  foot-pounds  in  1  hour. 
=  1,980,000,000  foot-pounds  in  10  hours. 
Each  gallon  requires  (8  x  200  =)  1,600  foot-pounds. 
1,980,000,000  -f-  1,600  =  1,237,500,  the  number  of  gallons. 

9.   Because  the  base  is  large,  and  the  center  of  mass  is  low. 

10.  Because  the  base  is  larger.     See  §  97. 

11.  One  is  at  the  center  of  the  axle,  and  the  other  should  be. 

12.  In  the  first  case,  the  center  of  mass  was  so  low  that  the 
line  of  direction  fell  within  the  base  ;  in  the  other  case,  it  was 
so  high  that  the  line  of  direction  fell  without  the  base. 

Page  1O4. 

1.  The  center  of  mass. 

2.  No;  it  is  about  midway  between  the  two  sides  of  the 
cardboard. 

3.  The  three  lines  should  intersect  at  a  common  point. 

4.  The  difference  represents  the  distance  that  the  center  of 
mass  must  be  raised  to  throw  the  line  of  direction  outside  the 
base.     The  products  represent  the  amount  of  work  that  must 
be  done  in  each  case  to  overturn  the  box.     Foot-pounds. 

5.  The  red  patch  will  circle  around  the  black  patch  as  the 
latter  moves  through  the  air. 


IN  AVERY'S   SCHOOL   PHYSICS. 


23 


Page  113. 

4.  l'  =  $  g(2  t-l)=  16.08  ft.  x  5  -  80.4  ft. 

5.  I  =  $gt2  =  16.08  ft.  x  100  =  1,608  ft. 

6.  I  =  $  gt2  =  16.08  ft.  x  i  =  4.02  ft. 

7.  I  =  %gt2  =  16.08  ft.  x  2.25  =  36.18  ft. 

8.  2,512ft. 

9.  I  =%gt2-,  787.92  =  16.08 Z2;  49  =  **;  7  =  t. 

10.  v=gt  =  32.16  ft.  x  7  =  225.12  ft. 

11.  v=gt  =  32.16  ft.  x  15.5  =  498.48  ft. 

12.  7J  oz.  +  7£  oz.  +  1  oz.  =  16  oz.,  the  total  weight  to  be 
moved.     To  move  this  load  we  have  the  weight  of  the  rider,  a 
force  of  1  oz.     This  force  can  give  to  an  ounce  of  matter  a 
velocity  of  (</=)32.16  ft.  per  second;  the  same  force  can  give 
to  16  oz.  of  matter  only  T^  of  this  velocity. 

32.16  ft.  -s- 16  =  2.01  ft. 

13.  I  =  ±  gt- ;  257.28  =  16.08 12 ;  4  =  t. 

The  ball  will  reach  the  ground  at  the  end  of  4  seconds. 
During  that  time  it  will  move  from  the  tower  (60  ft.  x  4=) 
240  ft.  See  §§  62  and  111. 

14.  Suppose  the  sinker  to  be  at  A.     Gravity  pulls  or  pushes 
it  toward  B.     The  current  pushes  it  toward  C.     The  resultant 


of  these  tends  to  move  it  to  D.     The  equilibrant  of  AD  is  AE, 
which  represents  the  tension  or  pull  of  the  line. 


24  KEY   TO   THE   PROBLEMS 

15.  v  =  gt;  80.4  =  32.16 1;  2.5  =  t. 

The  body  will  rise  for  2£  seconds.  At  the  time  specified,  it 
will  have  been  falling  £  second,  and  will  have  a  velocity  of 
16.08  ft. 

16.  Fl  represents  the  distance  that  gravity  will  move  the 
body  during  the  first  second.     Fa  represents  the  velocity  due 
to  the  horizontal  impelling   force,  or  the  distance  that  force 
would  move  it  in  the  first  second.     Aa  represents  the  amount 
of  deviation  from  a  horizontal  plane,  as  a  consequence  of  the 
pull  of  gravity  during  the  first  second.     It  is  equal  to  Fl.     Fc 
represents  the  horizontal  distance  the  projectile  will  move  in  3 
seconds.     It  is  3  times  the  initial  velocity.     Dd  represents  the 
total  pull  of  gravity  for  four  seconds  from  starting. 

17.  I  =  i  gt2  =  16.08  ft.  x  16  =  257.28  ft. 

This  is  the  distance  the  body  would  have  moved  in  the  given 
time  if  it  had  had  no  initial  velocity.  But  it  moved  (357.28 
—  257.28=)  100  feet  further  in  the  4  seconds.  The  initial 
velocity  must,  therefore,  have  been  (100  -5-  4  =)25  feet. 

18.  During  that  time,  gravity  alone  moved  it  2,512.5  ft.    The 
additional  force  moved  it  (35  x  12£  =)  437.5  ft.    Together  they 
moved  it  (2,512.5  +  437.5  =)  2,950  ft.    Its  final  velocity  due  to 
gravity  is  (32.16  x  12.5  =)402  ft.,  to  which  we  must  add  the 
initial  velocity,  making  a  velocity  of  437  ft. 

19.  (a)  3,216  -r-  32.16  =  100,  the  number  of  seconds. 

(b)  The  end  of  the  4th  second  of  the  ascent  corresponds  to 
the  end  of  the  96th  second  of  the  descent. 

v  =  gt  =  32.16  ft.  x  96  =  3,087.36  ft. 

Do  not  forget  that  this  result  disregards  the  resistance  of 
the  air;  that  it  is  true  only  for  a  freely  falling  (or  rising) 
body. 

(c)  The  end  of  the  7th  second  of  the  ascent  corresponds  to 
the  end  of  the  93d  second  of  the  descent, 

v  =  gt  =  32.16  ft.  x  93  =  2,990.88  ft. 


IN  AVERY'S   SCHOOL   PHYSICS.  25 

20.  The  ball  has  to  fall  from  a  height  of  257.28  ft.     Whether 
the  force  of  gravity  draws  it  vertically  downward  as  a  freely 
falling  body,  or  acts  with  the  projecting  force  to  produce  a 
curved  path,  it  will  reach  the  ground  in  the  same  time.     It 
would  fall  this  distance  in  what  time  ? 

I  =  $  gt2 ;  257.28  =  16.08  t2 ;  16  =  t2 ;  4  =  t. 
If  it  would  take  4  seconds  to  fall  from  the  highest  point 
reached,  it  would  require  4  seconds  for   it   to   rise   to   that 
height;    it  would  be  in  the  air  8  seconds.     During  each  of 
these  seconds  it  has  a  horizontal  motion  of  1,000  ft. 

21.  l=±gt*  =  16.08  ft.  x  25  =  402  ft.,  the  distance  that  the 
force  of  gravity  would  move  the  body  in  5  seconds.     The  force 
with  which  it  was  thrown  moves  it  10  ft.  each  second,  or  50 
ft.  during  the  five  seconds. 

402  ft.  +  50  ft.  =  452  ft. 

22.  v  =  gt  =  32.16  ft.  x  5  =  160.8  ft. 

160.8  ft.  + 10  ft.  =  170.8  ft. 

23.  See  Experiment  54.     (a)  The  value  of  each  space  is  7  ft. 
In  5  seconds  it  will  pass  over  (t2  =  )  25  such  spaces,  or  175  ft. 

(b)  Its  final  velocity  will  be  (2 1  =  )  10  times  7  ft.  =  70  ft. 

24.  (a)  See  §  107  (c).     The  distance  passed  over  in  the  first 
second  will  be  half  the  velocity  acquired  during  the  first  sec- 
ond.    Hence,  the  value  of  the  spaces  traversed  in  this  case  is 
10  ft.     In  10  seconds  it  will  pass  over  (t2  =)  100  such  spaces, 
or  1,000  ft.     Or,  we  may  say  that  the  increment  of  velocity  (g) 
under  these  circumstances  is  20  ft.,  instead  of  32.16  ft.     Then, 

I  =  i  gt2  =  10  ft.  x  100  =  1,000  ft. 

(b)  See  §  107  (a).     The  ratio  between  the  height  and  length 
of  the  plane  is  the  same  as  that  between  20  ft.  and  32.16  ft. 
20  :  32.16  :  :  800  ft.  :  1,286.40  ft. 

25.  (a)  It  would  take  just  as  long  to  rise  1,302.48  ft.  as  it 
would  to  fall  that  distance.     Then, 

l  =  tgP;  1,302.48  =  16.08 12 j  81  =  t2 ;  9  =  t. 


26  KEY   TO   THE   PROBLEMS 

(&)  The  initial  velocity  of  a  body  that  can  rise  for  9  seconds, 
is  the  same  as  the  final  velocity  of  a  body  that  has  fallen  for  9 
seconds.  Then, 

v  =  gt  =  32.16  ft.  x  9  =  289.44  ft. 

26.  During  7  seconds,  gravity  would  give  it  a  velocity  of 
(gt  =  32.16    ft.  x  7  =)   225.12    ft.      But   as    its   velocity  is 
235.12  ft.,  it  must  have  had  an   initial  velocity  of   (235.12 
—  225.12  =  )  10  ft.     During  4  seconds,  gravity  would  move  it 
(±gt2  =  16.08  ft.  x  16  =)  257.28  ft.     But  during  each  of  these 
4  seconds,  the  force  of  the  throw  moves  it  10  ft.  more.     This 
amounts  to  40  additional  feet  during  the  4  seconds. 

257.28  ft.  +  40  ft.  =  297.28  ft. 

27.  I  =  i  gt- ;  787.92  =  16.08  t* ;  49  =  P ;  7  =  t. 

This  is  the  time  that  the  second  body  was  in  falling  787.92 
ft.  But  the  first  body  fell  3  seconds  more,  or  10  seconds. 
During  10  seconds  it  would  fall  16.08  ft.  x  100  =  1,608  ft. 

28.  During  4  seconds  it  will  fall  257.28  ft.     During  6  sec- 
onds it  will  fall  578.88  ft.     During  the  5th  and  6th  seconds 
it  will  fall  578.88  ft.  -  257.28  ft.  =  321.6   ft.     Or  we  may 
say  that  its  average  velocity  during  these  two  seconds  will  be 
that  attained  at  the  end  of  the  5th  second,  which  is  160.8  ft. 
Moving    at    this   average   rate   for  2   seconds,   it  will   move 
(160.8  ft.  x  2  =)  321.6  ft. 

29.  Inversely ;  as  the  time  increases,  the  velocity  decreases. 

30.  When  they  have  been  extended  to  an  infinite  distance. 

Page  117. 

4.  For  the  sake  of  illustration,  suppose  that  the  origin  of 
co-ordinates  is  taken  at  a  corner  of  one  of  the  squares  near  the 
upper  left-hand  corner  of  the  sheet  of  cross-section  paper. 
Let  each  division  on  the  axes  represent  10  ft. ;  then  the  suc- 
cessive points  will  fall  on  successive  vertical  ruled  lines.  In 


IN   AVERY'S  SCHOOL  PHYSICS. 


27 


the  first  half  second,  the  ball  will  move  10  ft.  toward  the  right 
and  4.02  ft.  downward.  The  corresponding  point  will  be  one 
space  to  the  right  of  the  axis  of  ordinates  and  about  0.4  of  a 
space  below  the  axis  of  abscissas.  The  co-ordinates  for  the 
successive  periods  will  be  as  follows : 


SECONDS. 

ABSCISSAS. 

OKDINATES. 

SECONDS. 

ABSCISSAS. 

ORDINATES. 

0.5 

1 

-4.02 

3.0 

6 

-144.72 

1.0 

2 

-16.08 

3.5 

7 

-196.98 

1.5 

3 

-36.18 

4.0 

8 

-257.28 

2.0 

4 

-64.32 

4.5 

9 

-325.62 

2.5 

5 

-100.50 

5.0 

10 

-402.00 

The  mapping  of  the  curve  will  require  11  or  12  spaces  to  the 
right  of  the  axis  of  ordinates,  and  49  or  50  spaces  below  the 
axis  of  abscissas.  The  same  scale  being  used  for  both  ordi- 
nates, the  curve  will  not  be  distorted,  but  will  accurately  repre- 
sent the  actual  path  of  the  ball.  Find  the  point  on  the  curve 
that  has  an  abscissa  of  6.5 ;  i.e.,  a  point  that  is  6.5  spaces  to 
the  right  of  the  axis  of  ordinates.  This  point  corresponds  to 
a  period  of  3.25  seconds.  Measure  its  ordinate ;  i.e.,  the  num- 
ber of  spaces  that  it  is  below  the  axis  of  abscissas.  Remember 
that  the  scale  adopted  is  1  space  =  10  feet,  and  determine  the 
magnitude  represented  by  the  ordinate.  It  should  be  nearly 
170  feet.  Next,  find  the  point  on  the  curve  that  has  an  abscissa 
of  11 ;  i.e.,  a  point  that  is  11  spaces  to  the  right  of  the  axis  of 
ordinates.  This  point  corresponds  to  a  period  of  5.5  seconds. 
Measure  its  ordinate ;  i.e.,  the  number  of  spaces  that  it  is  below 
the  axis  of  abscissas.  Determine  the  magnitude  represented 
by  this  ordinate.  It  should  be  about  486  feet. 

5.  Divide  the  several  values  of  I  by  the  square  of  the  corre- 
sponding number  of  time  intervals.  For  instance,  divide  the 
value  of  I  as  shown  in  Fig.  64,  by  the  square  of  15.  The 
several  quotients  will  be  equal  to  the  value  of  /  when  t  =  1. 


28  KEY   TO   THE   PROBLEMS 

Page  124. 

NOTE.  —  Take  39.1  inches  or  99.33  centimeters  as  the  length  of  a 
seconds  pendulum. 

1.  The  period  is  3  seconds.     See  §  115  (3).     39.1  in.  x  32  = 
351.9  in.,  the  length  of  the  pendulum. 

2.  The  period  is  2  seconds.     39.1  in.  x  22  =  156.4  in. 

3.  We  have  two  pendulums  to  compare ;  the  seconds  pendu- 
lum, the  length  and  period  of  which  are  known,  and  the  given 
pendulum,  the  length  of  which  is  30  inches,  but  the  period  of 
which  we  are  to  find. 

39.1  in.  :  30  in.  : :  I2 :  £2;  *  =  0.87.  Since  the  period  is  0.87 
seconds,  there  will  be  as  many  oscillations  in  60  seconds  as  0.87 
is  contained  times  in  60,  which  is  68.9,  the  number  of  oscilla- 
tions per  minute. 

4.  39.1 : 16  =  I2 :  t2 ;  t  —  0.64  — .    The  number  of  vibrations 
is  93.7  +  . 

5.  60  -T-  \  =  240,  the  number  of  oscillations  per  minute. 
39.1 : 1 :  :  I2 :  (|)-;    I  =  2.44+   in.      Or,  we  may  say  that,  as 

the  time  is  J  that  of  the  seconds  pendulum,  the  length  will  be 
TL  that  of  the  seconds  pendulum,  or  2.44+  in. 

6.  60  sec.  -f-  15   sec.  =  4,   the   number   of    oscillations   per 
minute. 

39.1 : 1 :  :  I2 :  152.  Or,  since  the  time  is  15  times  that  of  the 
seconds  pendulum,  the  length  will  be  (15*=)  225  times  the 
length  of  the  seconds  pendulum. 

7.  39.1 :  39.37  : :  I2 :  t2 ;  t  =  1+  sec. 

Notice  how  closely  the  meter  corresponds  to  the  length  of 
the  seconds  pendulum. 

8.  The  period  is  6  seconds. 

39.1  in.  x  62  =  1,407.6  in.  =  117.3  ft. 

9.  39.1  in.  :  10  in.  :  :  I2 :  £2 ;    t  =  1+.      If   the    period   is  | 
second,  the  number  of  oscillations  per  second  is  2. 


IN  AVERY'S   SCHOOL   PHYSICS.  29 

10.  The  period  is  60  seconds.     39.1  in.  x  602  is  equivalent 
to  11,730  ft.,  or  more  than  2^-  miles. 

11.  The  length  of  this  pendulum  is  that  of  the  seconds  pen- 
dulum.    Hence  the  number  of  oscillations  is  1  per  second,  and 
the  period  is  1  second. 

12.  99.33  cm.  x  22  =  397.32  cm.  =  3.9732  m. 

13.  99.33  cm.  x  1202  =  1,430,352  cm.  =  14.3+  Km. 

14.  (b)  99.33  :  24.83  ::  I2  it2-, 


15.  The  period  is  £  second.     99.33  cm.  x  (|)2  =  1.552  cm. 

16.  99.33  :  397.32::  I2  :t2;      t  =  2. 

17.  99.33  :  11.03  :  :  I2  :  t2  ;        t  =  ±  nearly. 

18.  99.33  cm.  x  102  =  9,933  cm.  =  99.33  m. 

19.  99.33  :  2,483.25  :  :  I2  :  t2  ;  t  =  5. 
SUGGESTION.  —  Divide  the  1st  couplet  by  99.33. 

20.  (a)  99.33  cm.  x  42  =  1,589.28  cm.  =  15.8928  m. 

21.  See  §  115  (3).    pip':  :  V4:  V49;  i.e.,  the  periods  will 
be  as  2  :  7. 

22.  1:  I'::  702  :  802;  i.e.,  the  lengths  will  be  as  49  :  64. 

23.  The  period  of  the  longer  pendulum  will  be  twice  that 
of  the  shorter  one. 

24.  39.1  in.  x  (i)2  =  1.564  in. 

25.  (a)  39.1  in.  x  82  =  2,502.4  in.,  or  99.33  cm.  x  82 

=  6,357.12  cm.  =  6,357+  m. 
(b)  39.1  in.  x  (i)2  =  0.61  in.,  or  993.3  mm.  x  (i)2  =  15.5  mm. 

26.  39.1  in.  x  (3.5)2  =  478.975  in.,  nearly  40  ft. 

27.  Time  of  vibration  =  0.8  seconds. 

39.1  in.  x  (0.8)*  =  25.02+  in. 


30  KEY   TO   THE   PROBLEMS 

28.  60  :  601  :  :  w*  :  4002;  n  =  398+. 

29.  See  §  115. 

1  =  ir~l-L-  ;   I  =  99.396,  the  number  of  centimeters. 

30.  The  period  is  f|f  . 

300      150  83        1502     7r2x83 


Page  125. 

i.  a  =  j;  2/  =  i;  «  =  TV;  2V;  sV 

2.  Substitute  the  known  values  in  the  formula  given  in 
§  115.     t  =  1,  and  I  is  determined  by  measurement. 

3.  The  period  is  0.5  of  a  second,  or  half  as  long  as  that  of 
the  other  pendulum.     Its  length  should  be  one-fourth  as  great. 

4.  The  period  is  \  of  a  second. 

7.   The  upper  bullets  correspond  to  shorter  pendulums,  and 
tend  to  move  faster  than  the  lower  bullets. 


Page  134. 

1.  See  §  124. 

(a)   5Q  *  15  =  3,  the  number  of  feet. 

KA   v  1  K 

(&)   ru  x  ±0  =  10,  the  number  of  pounds. 
75 

2.  (a)    1QQ  x  1Q  =  8,  the  number  of  pounds. 


(6)    1QQ  x  10  =  5,  the  number  of  feet  per  second. 
200 

3.  The  lever  has  equal  arms;  i.e.,  it  is  a  balance.     The 
power  and  load  are  equal. 

4.  (a)  15  feet,     (b)  10  feet,      (c)  Such  a  lever  cannot  be 
of  the  3d  class,  for  in  such  levers  the  weight-arm  is  always 
greater  than  the  power-arm. 


IN  AVERY'S  SCHOOL  PHYSICS.  31 

5.  (a)  First  and  second  classes.     (6)  If  of  the  1st  class,  30 
Ibs. ;  if  of  the  2d  class,  40  Ibs. 

6.  Figure  the  lever.    It  must  be  of  the  1st  class,  with  arms 
of  4  and  6  feet.     See  §  131. 

40  x  4  =  160 ;  1,000  x  2  =  2,000 ;  160  +  2,000  =  2,160, 

the  sum  of  the  moments  for  the  short,  and.  consequently,  the 
moment  of  the  force  to  be  applied  at  the  end  of  the  long  arm 
to  produce  equilibrium.  But  in  the  case  of  the  long  arm,  6, 
the  length  of  the  arm,  is  one  factor  of  this  2,160 ;  there- 
fore (2,160-5-6=)  360,  the  number  of  pounds,  is  the  other 
factor.  Or,  more  briefly, 

6  x  =  2,160 ;     x  =  360. 

7.  The  lever  is  of  the  first  kind,  with  arms  of  8  and  12 
feet   respectively.      Figure   the   lever.      Eepresent  the  force 
at  the  end  of  the  short  arm  by  x.     Then  will  the  force  at  the 
end  of  the  other  arm  be  represented  by  1,200  —  x.     Placing 
the  moments  of  the  two  forces  equal  to  each  other,  we  have, 

8  x  =  12(1,200  -  a?)=  14,400  -  12  x. 
.-.  20  x  =  14,400;  a  =  720;  1,200  -  x  =  480. 

Ans.    720  Ibs.  at  the  end  of  the  short  arm,  and  480  Ibs.  at 
the  end  of  the  long  arm. 

Proof  {720  +  480  =  1,200. 

'(720x      8=    480x12. 

8.  Figure  the  lever,  with  the  first  three  forces  acting. 


40 
300 


100 


340  =  100  +  240. 

We  thus  see  that  an  additional  moment  of  240  is  needed 
with  the  100,  to  produce  equilibrium.  Of  this  240,  one  factor 
is  80 ;  then  the  other  factor  is  3,  the  number  of  feet  from  the 
fulcrum,  on  the  same  side  as  the  force  of  100  Ibs.  Or,  we  may 


32  KEY  TO   THE   PROBLEMS 

represent  the  four  forces  acting  upon  the  lever,  calling  the  dis- 
tance of  the  fourth  from  the  fulcrum,  x.    Then  we  shall  have 


40 
300 


100 

80  x 


340  =  100  +  80  x;    x  =  3. 

SUGGESTION. — In  a  case  like  this,  simple  inspection  will  generally 
show  on  which  arm  to  figure  the  fourth  force  as  acting.  If  it  is  errone- 
ously represented,  either  as  to  position  or  direction,  the  result  will  be  a 
negative  quantity.  For  instance,  suppose  that  in  this  case  we  represent 
the  force  of  80  Ibs.  as  acting  on  the  same  arm  as  the  forces  of  10  and  of 
100  Ibs.,  and  in  a  downward  direction.  We  should  then  have 


40 
300 

80  x 


100 


340  +  80  x  =  100  ;     80  x  =  -  240  ;     x  =  -  3. 

Here  the  3  would  indicate  that  the  force  was  acting  at  a  distance  of  3  ft. 
from  the  fulcrum ;  the  minus  sign  would  indicate  that  the  force  was 
tending  to  turn  the  lever  in  a  direction  opposite  to  that  assumed.  This 
contrary  tendency  might  result  from  an  upward  force  of  80  Ibs.  acting 
3  ft.  from  the  fulcrum  and  on  the  same  arm  as  the  10  Ibs.,  or  from  a 
downward  force  of  80  Ibs.  acting  3  ft.  from  the  fulcrum  and  on  the  same 
arm  as  the  100  Ibs.  Due  regard  for  the  algebraic  sign  of  the  result  will 
correct  any  error  of  this  kind.  In  this  problem,  a  downward  force  is 
specified ;  the  problem  is  therefore  definite. 

9.  Draw  a  line,  a&,  to  represent,  on  any  convenient  scale, 
as  1  inch  to  the  foot,  a  lever  6J-  feet  long.  Then  this  line 
will  be  61  inches  long.  Measure  on  the  line  f  of  an  inch  from 
b,  for  the  position  of  the  fulcrum,  c.  The  lever  is  evidently 
of  the  1st  class,  with  arms  of  5J  feet  and  f  feet  respectively. 
Represent  a  downward  force  of  60  pounds  at  a  (Fig.  76). 
Locate  d,  2|  inches  (feet)  from  c.  Represent  a  downward 
force  of  75  pounds  acting  at"  d.  Represent  a  downward  force 
of  x  pounds  at  b. 

51  x  60  =  330 

2J  x  75  =  2061 


536}  =  |  x;    z  = 


IN  AVERY'S  SCHOOL  PHYSICS.  33 

10.  Figure  the  lever  as  in  Exercise  9.  Place  the  moments 
tending  to  move  a  upward  in  one  column,  and  those  tending 
to  move  it  downward  in  the  other  column. 


40  x  10  =  400 

28  x 


364  =  56  x  6 
288  =  96  x  3 


28  x  +  400  =  652 

28  x  =  252 

x=      9 

The  force  of  28  Ibs.  must  act  at  a  distance  of  9  ft.  from  the 
fulcrum.  It  must,  therefore,  act  on  the  long  arm,  as  the  short 
arm  is  only  3  ft.  long.  We  assumed  that  the  force  would  tend 
to  draw  a  downward.  This  assumption  was  not  contradicted 
by  a  negative  result.  The  force  will  therefore  act  downward, 
at  a  distance  of  1  ft.  from  a. 

11.  Draw  a  line  18  cm.  long  to  represent  the  beam.  (Scale : 
1  cm.  =  1  ft.)  Letter  its  extremities  a  and  b.  At  a  point  3 
cm.  from  a,  represent  a  downward  force  of  2,000  Ibs.  At  a 
point  8  cm.  from  &,  represent  a  downward  force  of  1,400  Ibs. 
Consider  either  end  as  the  power,  and  the  other  end  as  the 
fulcrum. 

WITH  POWER  AT  b. 

2,000  x  3=  6,000 
1,400  x  10  =  14,000 


20,000    =18  a; 
1,1111="     x 

Pressure  at  b  =  IjH-H  Iks. 

Pressure  at  a  =  (3,400  -  1,111^  =)  2,288|  Ibs. 


WITH   POWER  AT  a. 

11,200  =  1,400  x    8 
x     30,000  =  2,000  x  15 


18  x  =  41,200 

x=    2;288f 

Pressure  at  a  =  2,288f  Ibs. 

Pressure  at  b  =  (3,400  -  2,288f  =)  1,111J-  Ibs. 
3 


34  KEY   TO   THE  PROBLEMS 

12.   Eepresent  the  power-arm  by  x  and  the  weight-arm  by 
3  —  x.     Call  40  the  power,  and  200  the  weight. 


P:  W:  :  WF:  PF,  or  40  :  200  :  :  3  -  x  :  x. 

.-.  40 x  =  600  -200 x;  240^=600; 
a  =  21;  3-ic  =  f 

The  fulcrum  must  be  2^  ft.  from  the  40  Ibs.,  and  6  in.  from 
the  200  Ibs. 

Or,  we  may  proceed  as  follows :  The  weight  is  5  times  the 
power;  consequently,  the  power-arm  must  be  5  times  the 
weight-arm.  Dividing  36  inches,  the  length  of  the  lever, 
into  two  parts,  so  that  one  shall  be  5  times  as  great  as  the 
other,  we  have  30  inches  and  6  inches. 

13.  See  §   132  (a).     Eepresent  the  true  weight  by  x,  and 
remember  that  1  Ib.  9  oz.  =  25  oz.,  and  that  2  Ibs.  4  oz.  =  36  oz. 

25  :x::x:  36 ;  x2  =  900 ;  x  =  30. 
The  true  weight  is  30  oz.,  or  1  Ib.  14  oz. 

14.  A  ought  to  carry  200  Ibs. ;  B,  100  Ibs.     Call  the  6  ft. 
bar  a  lever  of  the  second  class,  fulcrum  at  either  end,  as  A's 
end.     Then  the  weight  is  300  Ibs. ;  the  power,  100  Ibs. ;  the 
power-arm,  6  ft. ;  the  weight-arm  =  x. 


P:W::WF:PF;  100  :  300  :  :  x  :  6; 
300^  =  600;  x  =  2. 

The  weight  should  be  hung  two  feet  from  the  fulcrum ;  or  two 
feet  from  A,  and  four  feet  from  B. 

15.  As  the  beef  weighs  450  pounds,  A  should  carry  250 
pounds,  and  B,  200  pounds.  Suppose  B  to  act  as  fulcrum, 
and  A  as  power. 

P:W::WF:PF;  250  :  450  : :  x :  8; 
9a  =  40;  #  =  4f 

The  beef  should  hang  4f  ft.  from  the  fulcrum ;  or  4J  ft.  from 
B,  and  3|  ft.  from  A. 


IN  AVERY'S  SCHOOL  PHYSICS.  35 

16.  The  lever  is  of  the  1st  class,  with  arms  of  3J  ft.  and 
(16  _  31  =)  121  ft.  respectively. 

100  x  121  =1,250  =  3Ja;   x  =  357f 

It  would  require  a  power  of  357-f  Ibs.  to  balance  the  100 
Ibs.  On  the  supposition  that  the  machine  is  free  from  fric- 
tion and  other  hindrances  to  motion,  anything  more  than 
357^-  Ibs.  would  move  the  load.  In  practice,  an  allowance, 
determined  by  experience,  is  made  for  these  necessary  hin- 
drances. 

17.  (a) 


(6)  50  x  40  x  33  =  x  x  6  x  5  ;  x  =  2,200,  the  number  of 
pounds. 

(c)  x  x  40  x  33  =  4,400  x  6  x  5  ;  x  =  100,  the  number  of  Kg. 

SUGGESTION.  —  Theoretically,  the  parts  of  a  lever  or  other  simple 
machine  have  no  weight,  suffer  no  loss  by  friction,  from  the  stiffness  of 
ropes,  the  resistance  of  the  air,  or  similar  causes.  In  practical  mechanics, 
these  have  to  be  taken  into  consideration. 

18.   The  weight  of  the  bar,  4  Ibs.,  may  be  considered  as  con- 
centrated at  the  center  of  mass,  5  in.  from  either  end.     See 
§  94  (a).     Let  x  represent  the  distance  of  the  fulcrum  from 
the  end  at  which  the  6-lb.  weight  is  hung. 
6x  =  4(5-z);  a?  =  2. 

Figure  the  lever,  and  notice  the  equality   of   the   moments; 
3x4  =  2x6. 

Page  136. 

2.   The  reading  of  the  dynamometer  includes  the  weight  of 
the  bar. 

10.  In  every  case,  the  weight  of  the  rod  acts  as  if  it  was 
concentrated  at  the  same  point,  viz.,  at  the  center  of  mass. 


36  KEY   TO   THE   PROBLEMS 

Page  14O. 

1.  P:  W::  d:  D;  x  :  180  ::  6:  36;  x  =  30. 
Any  power  greater  than  30  Ibs.  will  move  the  rudder. 

2.  x  :  2,000  : :  8  :  80  ;  x  =  200. 
200  -r-  4  =  50  Ibs.     Ans.  50+  Ibs. 

3.  See  Fig.  85.     P:  W: :  d  :  D;  x:  1,100  ::  1 : 10;  x  =  110; 

110  -7-  4  =  27-J-,  the  number  of  pounds. 

4.  The  circumference  of  the  axle  is  ^  of  the  circumference 
of  the  wheel ;  1  of  85  Ibs.  is  14J  Ibs. 

5.  70  :  300  : :  x  :  120 ;  x  =  28. 

Or,  the  power  is  -j^  of  the  weight ;  hence,  the  diameter  of  the 
axle  must  be  -^  of  120  inches  —  28  inches. 

6.  The  diameter  of  the  circle  described  by  the  power  is  36 
inches. 

(a)    x  :  62.5  : :  10  :  36;    x=  17.36  +  ,  the  number  of  pounds. 
(6)   20  liters  of  water  weigh  20  kilograms. 

x  :  20  : :  10  :  36 ;  x  =  5.55,  the  number  of  Kg. 

7.  We  have  two  sets  of  twro  men  each.     The  first  set  move 
in  a  circle  14  feet  in  diameter ;  the  second  set  in  a  circle  10 
feet  in  diameter.     Find  the  effect  produced  by  each  set  and 
add. 

1st  set 60  :  x   : :  14  :  12  x  14 ;  x  =     720 

2dset 80:  a;':;  14;  12x10;  x'  =     685^ 

Total  effect  =  1,405  Ibs. 

8.  The  diameter  of  the  circle  traversed   by  the   horse   is 
14  feet.     The  diameter  of  the  capstan  barrel  is  14  inches ;  i.e., 
the  first  diameter  is  12  times  the  second  diameter.     Hence,  the 
circumference  of  the  circle  traversed  by  the  horse  is  12  times 
as  great  as  the  circumference  of  the  barrel  of  the  capstan. 
From  this  it  follows  that  the  horse  must  move  12  times  as  far 
as  the  house.     (This  may  be  made  clear  by  computing  the  two 


IN  AVERY'S   SCHOOL   PHYSICS. 


37 


circumferences ;  one  of  them  will  show  the  distance  the  horse 
travels  in  turning  the  barrel  around  once ;  the  other  will  show 
how  much  rope  was  wound  up,  or  how  far  the  house  moved  for 
that  revolution  of  the  capstan  barrel.)  Then  the  horse  must 
travel  (5  miles  x  12  =)  60  miles.  At  the  given  rate,  this  will 
require  (60  -f-  2|  =)  24  hours. 

Page  145. 

1.  50  pounds.     The  fixed  pulley  gives  no  mechanical  advan- 
tage.    See  §  139. 

2.  It  may  be  25  Ibs.  (see  Fig.  88),  or  16 1  Ibs.  (see  Fig.  89). 

3.  75  Ibs.  x  7  —  525  Ibs.      The  friction,   etc.,  help  to  sup- 
port the  load,  so  that  the  power  may  be  considered  as  90  Ibs. 
90  Ibs.  x  7  =  630  Ibs. 

4.  2,000  -r-  10  =  200,  the  number  of  pounds. 

5.  2,000 -r- 11  =  181T9T+,  the          7. 
number  of  pounds.     Friction 
resists   the  motion,   and  may 

be  considered  as  added  to  the 
load.  2,500  Ibs.  -=-  11  =  227y\, 
the  number  of  pounds. 

6.  There  may  be  12  cords 
or  13  cords.    The  friction  will 
support  \  of  the  weight,  leav- 
ing 1,350  Ibs.  to  be  otherwise 
supported. 

1,350  Ibs.  -5- 13  =  10311  Ibs. 
The  power  will  move  13  times 
as  fast  as  the  weight,  if  either 
moves  at  all. 

8.  The  center  of  mass  of  the  125-pound  board  is  10  feet 
from  each  end  of  the  board.  When  the  boy  lifts  at  one  end  of 
the  board,  the  other  end  becomes  the  fulcrum  of  a  lever  of  the 


38  KEY  TO   THE   PROBLEMS 

second  class,  and  the  power  required  is  only  half  the  load  to 
be  raised.  He  is,  therefore,  strong  enough  to  lift  one  end  of 
the  plank  as  required. 

20  :  5  : :  196  :  x;  x  =  49. 

9.  It  involves  just  as  much  work  and  requires  as  much 
energy  to  move  it  up  the  inclined  plane  as  it  would  to  lift  it 
outright,  namely,  600  foot-pounds.  This  is  involved  in  the 
very  definitions  of  work  and  energy,  but  it  may  be  well  to 
illustrate  it.  The  power  required  to  support  the  weight  on 
the  plane  (ignoring  friction)  is  determined  according  to  the 
law  given  in  §  141  (1) : 

10  ft.  :  6  ft.  =  100  Ibs.  :  60  Ibs. 

But  this  power  of  60  pounds  must  act  along  the  whole  length 
of  the  plank,  i.e.,  through  a  distance  of  10  feet,  in  order  to 
lift  the  weight  the  required  distance. 

60  x  10  =  100  x  6. 

The  result  is  600  foot-pounds,  whether  it  is  reached  in  one  way 
or  in  the  other.  But  the  use  of  the  inclined  plane  introduces 
friction  and  increases  the  work  to  be  done  by  one-fifth,  or  120 
foot-pounds,  making  a  total  of  720  foot-pounds. 

10.  4,000  Ibs.  X  T3o  -  1,200  Ibs. ;  or  P :  W:  :  h  :  b; 

P:  4,000::  3:  10;  P=  1,200. 

11.  (a)  If  the  power  acts  horizontally, 

50  Ibs.  x  LO  =  125  Ibs.     Ans.  Nearly  125  Ibs. 

(6)  If  the  power  acts  in  the  direction  of  the  plane,  first  find 
the  length  of  the  plane  : 

yi02  +  42  =  10.77  +  . 
Then,  P:  W: :  h  :  I;     W=  nearly  134.625  Ibs. 

12.  1  of  40  Ibs.  is  5  Ibs. 

13.  The  power  is  (g2-^  =)  -fa  of  the  weight.     Consequently, 
the  base  is  32  times  as  long  as  the  height,  or  256  ft.     The 


IN   AVERY'S  SCHOOL  PHYSICS.  39 

length  is  the  hypotenuse  of  a  right-angled  triangle,  having 
sides  of  8  and  256  ft.  respectively. 


V82  +  2562  =  256.1. 

SUGGESTION.  —  If  you  have  pupils  in  the  class  who  can  do  original 
work,  let  them  try  to  solve  the  problem  by  construction. 

14.  75  Ibs. :  4,000  Ibs.  =  1  yd. :  531  yds.,  or 
75  Ibs.  :  4,000  Ibs.  =  3  ft. :  160  ft. 

15.  PiWiih:  1;     250  : 1,500  :  :  5  :  Z;     I  =  30  ft.  or  more. 

16.  The  load  is  400  Ibs.  +  10  Ibs.  =  410  Ibs.     The  movable 
pulley  divides  the  rope  into  4  parts,  and  the  load  is,  therefore, 
4_p_  =  102 1.  Ibs.     The  support  of  the  upper  block  has  to  carry 
both  blocks,  the  power,  and  the  load,  or  20  Ibs.  + 1021  Ibs.  -f 
400  Ibs.  =  5221  ibs. 

Page  148. 

1.  8  ft.  =  96  in.     96  in.  -5- 1  in.  =  288.     The  power  moves 
96  in.  while  the  weight  moves  J  in.,  or  it  moves  288  times  as 
fast.    Hence,  the  theoretical  pressure  will  be  288  times  15  Ibs.  or 
4,320  Ibs.    Deducting  for  friction,  we  have  4,320  Ibs.  -  240  Ibs. 
=  4,080  Ibs.,  the  actual  pressure. 

2.  Diameter  =  16  in. ;    circumference  =  16  in.  x  3.1416  = 

50.2656  in. 

50  Ibs.  x  50.2656  x  11  =  27,646.08  Ibs. 

Deduct  i  of  this  for  friction,        3,455.76  Ibs. 
Available  power 24,190.32  Ibs. 

Any  resistance  less  than  24,190.32  Ibs.  may  be  overcome.  The 
friction  may  be  provided  for  by  deducting  1  of  the  50  Ibs.  before 
multiplying  by  50.2656  x  11.  The  problem  may  also  be  solved 

as  follows : 

W  X  TV  =  50  Ibs.  x  50  x  50.2656.       See  §  124  (1). 

SUGGESTION.  — The  distances  here  used  are  those  traversed  during  one 
revolution  of  the  screw. 


40  KEY   TO   THE   PROBLEMS 

4^x  2x12  x  3.1416  =  2nA  + 

Of  course  this  ignores  friction. 

4.   15  Ibs.  x  8  -  120  Ibs. 

P:W::r:E;       25  :  W :  :  41 :  30;       W=  166|. 

The  boy  could  support  a  load  of  166|  Ibs.  with  the  wheel 
and  axle,  without  the  aid  of  the  inclined  plane.  With  the 
inclined  plane,  he  could  support  20  times  as  great  a  load,  or 
3,3331  Ibs.  He  could  lift  anything  less  than  3,333J  Ibs. 

6.  5001bs.x7^x2xl2x6  =  ^^  ibs. 

Deducting  \  of  this  for  loss  by  friction  .     11,454.54  Ibs. 
we  have,  for  the  force  exerted 34,363.64  Ibs. 

Page  149. 

3.  P:TF::1:4. 

4.  There   is   no  difference   in   the   work   performed.      The 
power  is  \  as  large,  but  it  moves  4  times  as  far. 

5.  The  ratio  is  16.      In   general  terms,  P :  W :  :  1  :  2%  or 
W—Px  2n,  in  which  n  represents  the   number  of  movable 
pulleys. 

6.  Construct  the  figure  very  carefully.      The  larger  it  is, 

the  better.  Represent  the  base, 
AB,  by  a  line  at  least  20  cm. 
long,  and  make  AC  one-fifth  as 
long.  Of  course,  AB  is  supposed 
to  be  horizontal,  so  that  the  angle 
B  measures  the  inclination  of  the 
plane  with  reference  to  the  hori- 
zon. Anywhere  on  BC,  take  the 
point  W  to  represent  the  position 
of  the  globe.  Represent  the  grav- 
ity of  the  globe  by  a  vertical  line.  If  you  adopt  the  scale  of 
1  mm.  =  2  Kgv  this  line,  WG,  will  be  12.5  cm.  long.  From  W, 


IN   AVERY'S   SCHOOL   PHYSICS. 


41 


draw  WX,  so  that  it  shall  make  an  angle  of  45°  with  AB. 
This  may  be  done  by  taking  oi  =  o  W,  and  drawing  a  line 
from  W  through  i.  From  W,  draw  WZ  perpendicular  to  BG. 
From  G,  draw  GP  parallel  to  WZ,  and  GE  parallel  to  WP. 
WE  represents  the  pressure  of  the  globe  upon  the  plane,  and 
WP  the  tendency  to  move  in  the  direction  WX.  This  ten- 
dency must  be  resisted  by  the  supporting  force  which  equals 
it  in  intensity  and  is  opposite  to  it  in  direction.  PW  will 
measure  30+  mm.,  and  represent  a  force  of  60+  Kg. 

7.  Draw  ABC,  WG,  and  WZ,  as  in  the  last  problem. 
From  W,  draw  WX'  perpendicular  to  WX.  From  G,  draw 
a  line  parallel  to  WZ,  and  prolong  it  until  it  intersects  WX' 
at  P'.  Complete  the  parallelogram  by  drawing  GE1  parallel  to 
WP'.  Measure  GP',  and  find  its  value  according  to  the  scale 
adopted. 

SUGGESTION.  —  In  the  6th  problem,  the  supporting  force  tends  to  lift 
the  globe  from  the  plane,  and  thus  to  make  its  pressure  on  the  plane  less 
than  its  weight ;  in  the  7th,  the  sup- 
porting force  tends  to  press  the  globe 
against  the  plane,  and  thus  to  make 
its  pressure  on  the  plane  greater 
than  its  weight. 

9.  Draw  the  right-angled 
triangle,  ABC,  with  its  sides 
as  3  :  4  :  5,  as  shown  in  the  ac- 
companying figure.  Then  will 
AC  represent  the  base,  and 
BC  the  length  of  the  plane. 
From  W,  the  point  where  the 
ball  rests  on  the  plane,  draw 
the  vertical  line,  WG,  1\  units 
long  to  represent  the  weight 
of  the  ball.  From  W,  draw  a 
line  perpendicular  to  BC,  and 
from  G  draw  a  line  parallel  to  BC,  so  that  they  will  intersect 
at  P.  Complete  the  parallelogram  WPGE.  WE  represents 


42  KEY  TO   THE   PROBLEMS 

the  tendency  of  the  ball  to  roll  down  the  plane,  and  WP  repre- 
sents the  pressure  of  the  ball  on  the  plane.  As  the  triangle, 
WPG,  is  similar  to  the  triangle  BAG,  it  follows  that 


PG:WG::3:5-y 
PW:  TF#::4:5;  PTF:71::4:5;  PW=  6. 

But  PG  (or  its  equal,  WE)  represents  the  tendency  of  the  ball 
to  roll  down  the  plane,  and  its  value  should  be  indicated  by 
the  reading  of  the  dynamometer  that  resists  that  tendency. 
Similarly,  PW  represents  the  pressure  of  the  ball  on  the  plane, 
and  its  value  should  be  indicated  by  the  reading  of  the  dyna- 
mometer that  measures  that  pressure. 

Page  158. 

1.  See  §  151  (2).     The  imaginary  column   has   a   base   of 
30  x  20  and  an  altitude  of  10  ft.     Cubic  contents  =  6,000  cu. 
ft.     The  weight  of  6,000  cu.  ft.  of  water  is  62.42  Ibs.  x  6,000 
=  374,520  Ibs. 

SUGGESTION.  —  Allow  the  pupil  to  use  62  1  Ibs.  as  the  weight  of  a  cu.  ft. 
of  water,  if  he  prefers  to  do  so.  That  value  is  more  easily  remembered, 
and  nearly  enough  accurate  for  non-professional  purposes. 

2.  Bulk  of  imaginary  column  =  (6  x  10  x  3  =)  180  cu.  m. 
=  180  Kl.  =  180,000  liters.     §  21.     Each  liter  of  water  weighs 
1  Kg.     Hence  the  pressure  is  180,000  Kg. 

3.  5  x  12  x  6  =  360,  the  number  of  cu.  ft. 
62.42  Ibs.  x  360  =  22,471.2  Ibs. 

4.  2x4x2  =  16,  the  number  of  cu.  m. 

16  cu.  m.  =  16,000  1.,  which  weigh  16,000  Kg. 

5.  2  cu.  yds.  =  54  cu.  ft. 

62.42  Ibs.  x  54  =  3,370.68  Ibs. 

6.  2  cu.  m.  =  2,000  1. 

2,000  1.  of  water  weigh  2,000  Kg. 

7.  25  cu.  ft.  of  water  weigh  (62.42  Ibs.  x  25  =)  1,560.5  Ibs. 


IN  AVERY'S   SCHOOL   PHYSICS.  43 

8.  30  x  30  x  800  =  720,000,  the  number  of  cu.  cm.     This 
quantity  of  water  weighs  720,000  g.  or  720  Kg. 

9.  3  x  3  x  7  =  63,  the  number  of  cu.  ft. 
62.42  Ibs.  x  63  =  3,932.46  Ibs. 

10.  The  dimensions  of  the  vessel  are  12  x  12  x  12.     The 
acid  stands  8  in.  deep.     12  x  8  x  4  =  384,  the  number  of  cu. 
in.  in  the  imaginary  column.     This  is  |  of  a  cu.  ft.     If  the 
imaginary  column  were  of  water,  it  would  weigh  -|  of  62.42  Ibs. 
=  13.871+  Ibs.     Such  a  column  of  acid  would  weigh  1.8  times 
13.871+  Ibs.  =  24.9678+  Ibs. 

11.  237  x  35  =  8,295,  the  number  of  cu.  cm.,  and  consequently 
the  number  of  grams. 

12.  237  x  35  =  8,295,  the  number  of  cu.  in. 
62.42  Ibs.  x  ff f|  =  299.7  Ibs. 

13.  The  water  must  stand  (-2¥7-=)4i  ft.  deep.     The  sides 
subjected  to  lateral  pressure  have  an  area  of  (10  x  4-J-=)  45 
sq.  ft.     45  x  21  =  101 J,  the  number  of  cu.  ft.  in  the  imaginary 
column  producing  lateral  pressure.     There  are  27  cu.  ft.  (2x3 
X  41)    in   the    column   producing    pressure    on    the    bottom. 
101.25  +  27  =  128.25. 

62.42  Ibs.  x  128.25  =  7,999.365  Ibs. 

14.  If  the  lever  of  the  press  is  of  the  second  class,  as  shown 
in  Fig.  108,  the  piston  of  the  pump  will  go  down  with  a  force 
of  (75  Ibs.  x  6  =)  450  Ibs.      The  area  of  the  cylinder  being 
200  times  as  great  as  that  of  the  piston,  the  weight  will  be  200 
times  450  Ibs.,  or  45  tons.     If  the"  lever  is  of  the  first  class,  the 
weight  will  be  (75  Ibs.  x  5  x  200  =)  75,000  Ibs.  =  37£  tons. 

15.  2  x  3  x  5  =  30,  the  number  of  cu.  ft. 
62.42  Ibs.  x  30  =  1,872.6  Ibs. 

16.  See  Appendix. 

Area  =  *&  =  3.1416  x  100  =  314.16. 
314.16  x  48  =  15,079.68,  the  number  of  cu.  in. 
15,079.68  cu.  in.  =  8.72  cu.  ft. 
62.42  Ibs.  x  8.72  =  544.3+  Ibs. 


44  KEY  TO   THE   PROBLEMS 

17.  1,000  cm.,  or  10  m. 

18.  Draw   a   sectional   view   of  the  vessel  and  tube.     The 
water  in  the  tube  is.  35  cm.  higher  than  the  upper  end  of  the 
cylinder;    the   pressure   is   35  grains  per  square  centimeter. 
The  water  in  the  tube  is  65  cm.  higher  than  the  lower  end 
of  the  cylinder;  the  pressure  is  65  grams  per  square  centi- 
meter. 

19.  Draw  a  sectional  view. 

(a)  The  jar  contains  2,000  cu.  cm.,  and  the  tube  30  cu.  cm. 
The  2,030  cu.  cm.  of  water  weigh  2,030  grams. 

(6)  The  base  is  50  cm.  below  the  surface  of  the  water,  and 
the  pressure  is  50  grams  per  square  centimeter, 
(c  and  d)   The  pressure  on  the  base  is  uniform. 

20.  (a)  The  surface  described  is  30  cm.  below  the  surface 
of  the  water,  and  the  pressure  is  30  grams  per  square  centi- 
meter. 

(b)  The  pressure  on  the  cover  is  uniform. 

(c)  99  x  30  =  2,970,  the  number  of  cubic  centimeters  in  the 
imaginary  column,-  and  the  number  of  grams  pressure. 

21.  The  pressure  on  the  base  is    (100  x  50  =)  5,000  grams. 
The  pressure  on  the  cover  is  (  99  x  30  =)  2,970  grams. 

The  weight  of  water  is  2,030  grams. 

Page  16O. 

3.  The    difference    in   the   mercury   levels   measures   the 
upward  pressures  at  the  several  depths  of  the  water  levels  in 
the  tube  below  the  water  levels  outside  the  tube.     The  upward 
pressure  varies  as  the  depth. 

4.  If  the  line  is  straight,  it  shows  that  m  and  w  vary  alike, 
i.e.,  in  the  same  ratio.     If  the  line  is  curved,  it  shows  that  one 
varies  more  rapidly  than  the  other. 


IN  AVERY'S   SCHOOL  PHYSICS.  45 


Page  166. 

1.  It  will  lose  the  weight  of  1  cu.  dm.  of  water.     §  153. 
This  is  the  weight  of  a  liter  of  water,  1,000  g.  or  1  Kg. 

2.  It  would  lose  the  weight  of  1  cu.  dm.,  or  1  liter  of  the 
liquid   (mercury),  which  would  weigh  13.6  times  1  Kg.,  or 
13.6  Kg.  =  13,600  g. 

3.  The  iron  is  supposed  to  be  immersed  in  the   mercury. 
The  mercury  pushes  it  up  with  a  force  of  13,600  g.     Gravity 
pulls  it  down  with  a  force  of  7,780  g.     It  must  be  held  down 
by  an  additional  force  of  5,820  g.     It  can  carry  a  load  of  5,820  g. 
If  the  iron  is  allowed  to  float  on  the  mercury,  it  will  displace 
only  7,780  g.  of  mercury,  thus  losing  its  own  weight.     §  154. 

4.  The  weight  of  a  cu.  ft.  of  water,  or  62.42  Ibs. 

5.  It  will   displace   100  cu.   cm.  of  water.      It  will  there- 
fore lose  100  g.  in  weight.      The  remaining  weight  will  be 
1,035  g. 

6.  It  loses  10  g.  in  weight  when  placed  in  water.     It  there- 
fore displaces  10  g.  or  10  cu.  cm.  of  water;   its  own  bulk  is 
10  cu.  cm. 

7.  Draw  a  sectional  view. 

(a)   The  cube  contains  (20  x  20  x  20  =)  8,000  cu.  cm. 
The  tube  contains  (2  x  .2  x  10  =)  40  cu.  cm. 
The  8,040  cu.  cm.  of  water  weigh  8,040  g. 

(5)  The  imaginary  column  (see  §  151)  contains  (20  x  20  x 
30=)  12,000  cu.  cm.,  and  weighs  12,000  g.  The  downward 
pressure  is  12,000  g. 

(c)  The  top  of  the  vessel  that  is  subjected  to  an  upward 
pressure  is  (400  —  4  =)  396  sq.  cm.  The  imaginary  column 
measures  3,960  cu.  cm.  The  upward  pressure  is  3,960  g.  The 
downward  pressure  minus  the  iipward  pressure  represents  the 
net  downward  tendency  or  weight.  1 2,000  g.  -  3,960  g.  =  8,040  g. 


46  KEY  TO   THE  PROBLEMS 

8.  In  Fig.  117,  the  weight  of  the  particle  is  represented  by 
B  W.  This  force  is  resolved  into  two  components,  of  which  BE 
is  perpendicular  to  the  surface  of  the  water.  This  component 
is  met  by  the  resistance  of  the  water.  The  other  component, 
BD,  pulls  the  water  particle  to  a  lower  level.  When  the  sur- 
face becomes  level,  BE  coincides  with  BW,  and  BD  vanishes. 

Page  173. 

1.  As  it  displaces  only  half  of  its  own  volume  of  water,  it 
weighs  only  half  as  much  as  its  own  volume  of  water  (see  §  154). 
Consequently,  its  density  is  one-half,  or  0.5. 

2.  The  volume  of   a  sphere  =  1  TrD3  (see   Appendix  1)  = 
8,181.2  cu.  cm.     This  quantity  of  water  would  weigh  8,181.2  g. 
This  quantity  of  aluminium  would  weigh  2.6  times  as  much,  or 
21,271.12  g.  =21.27+  Kg. 

3.  52.35  g.  -r-  5  g.  =  10.47,  the  density  of  silver. 

4.  (a)   695  g.  -j-  83  g.  =  8.37+,  the  density  of  brass. 
(6)   83  g.  x  0.792  =  65.736  g. 

695  g.  -  65.736  g.  =  629.264  g. 

5.  708  gr.  -=- 1,000  gr.  =  0.708,  the  density  of  the  benzoline. 

6.  2.4554  oz.  -  2.0778  oz.  =  0.3776  oz. 
2.4554  oz.  -r-  0.3776  oz.  =  6.5+. 

7.  4.6764  oz.  -r-  0.2447  oz.  =  19.11,  the  density  of  the  gold. 

8.  See  §  160  (2).     970  gr.  -  895  gr.  =  75  gr. 
970  gr.  -  910  gr.  =  60  gr. 

60  gr.  -f-  75  gr.  ==  0.8. 

9.  23  gr.  -r-  25  gr.  =  0.92,  the  density  of  the  oil. 

19  gr.  -=-  25  gr.  =  0.76,  the  density  of  the  alcohol. 

10.  1,536  g.  -1,283  g.  =  253  g. 
1,536  g.  -f-  253  g.  =  6.07. 

11.  Subtract  the  weight  of  bottle  from  the  other  two  weights. 
4.2544  g.  -*•  4.1417  g.  =  1.027 +  . 


IN   AVERY'S   SCHOOL  PHYSICS.  47 

12.    (1)  Weight  of  wood  and  sinker  in  air,  14  g. 

(2)  «        "  «  "          water,     8.5  g. 

(3)  "        "  water  displaced  by  both,        5.5  g. 

(4)  «        "  "  "          sinker 

(10  g.  ---10.5=) .954  g. 

(5)  Weight  of  water  displaced  by  wood,     4.546  g. 

(6)  Density  of  the  wood  (4  g.  -=-  4.546  g.  =)  0.897. 

.13.    Density  of  the  metal  is  1.73  ;  of  the  unknown  liquid,  0.67. 

14.  2,160  gr.  -  1,511.5  gr.  =  648.5  gr. 
2,160  gr.  -*-  648.5  gr.  =  3.33  +  . 

15.  (1)   Weight  of  ice  and  lead  in  air     .     .     24  Ibs. 

(2)  «        "  "          "       water     .     13.712  Ibs. 

(3)  "        "  water   displaced  by  ice 

and  lead 10.288  Ibs. 

(4)  Weight  of  water  displaced  by  lead,       1.4      Ibs. 

(5)  "        "  «  «       ice  .       8.888  Ibs. 

(6)  Density  of  ice  (8  Ibs.  -f-  8.888  lbs.  =  )     0.9+. 

16.  (a)   600  g.  -  545  g.  =  55  g. 

600  g. -55  g.  =  10.9+. 
(6)   600  g.  -  557  g.  =  43  g. 

43  g.  -55  g.  =0.78  +  . 
(c)    600  cu.  cm.  -f-  10.9  =  55+  cu.  cm. 

17.  The  simple  question  is,  "  How  much   does  that  stone 
weigh  in  water  ?  " 

§156     2.5  =  — — —    .-.  w'  =  1801bs. 
300  -  w' 

18.  An  equal  bulk  of  water  weighs  1,000  g. 

870  g.  -=- 1,000  g.  =  0.87,  the  density  of  the  turpentine. 

19.  The    volume    of    the   fragments    was   1,000   cu.  cm.  - 
675  cu.  cm.  =  325  cu.  cm.     The  fragments  weighed  1,487.5  g. 
—  675  g.  =  812.5  g.     An  equal   bulk  of  water   would   weigh 

325  g. 

812.5  g,  —  325  g.  =  2.5,  the  density  of  the  mineral. 


48  KEY  TO   THE   PROBLEMS 

20.  The  800  cu.  cm.  of  water  weigh  800  g.     The  200  cu.  cm. 
of  sand  weigh  1,350  g.  -  800  g.  =  550  g.     An  equal  bulk  of 
water  weighs  200  g.    550  g.  --  200  g.  =  2.75,  the  density  of  the 
sand. 

21.  1?OOQ  ~  4Q  =  8,  the  density  of  the  brass.      §  157. 


22.  »        +    '       =1.8,  the  density  of  the  acid.   See  §  160  (3). 
ZjUUU  -f~  ijuuu 

23.  The  given   body  weighs   10  g.     It   displaces  2.5  g.  of 
water  (an  equal  bulk  of  water  weighs  2.5  g.). 

10  g.  -r-  2.5  g.  =  4,  the  density. 

24.  The  coal  would  weigh  2.4  times  as  much  as  a  cubic  foot 
of  water  or  (62.5  Ibs.  x  2.4  =)  150  Ibs.     It  would   displace 
1  cu.  ft.  of  the  solution,  which  would  weigh  (62.5  Ibs.  x  1.2  —  ) 
75  Ibs.     It  will  lose  75  Ibs.  weight  when  in  the  saline  solution. 
150  Ibs.  -  75  Ibs.  =  75  Ibs. 

Or,  we  may  say  that  the  coal  will  weigh  as  much  as  2.4  cu.  ft. 
of  water,  and  that  the  solution  displaced  by  it  will  weigh  as 
much  as  1.2  cu.  ft.  of  water.  The  weight  less  the  loss  by 
buoyancy  will  be  the  weight  of  (2.4  cu.  ft.  —  1.2  cu.  ft.  =) 
1.2  cu.  ft.  of  water,  or  75  Ibs. 

A  comparison  of  the  given  densities  shows  that  the  coal  is 
twice  as  heavy  as  the  solution.  Under  such  conditions,  the 
weight  of  the  coal  in  the  solution  will  equal  the  weight  of  the 
solution  displaced. 

25.  With  a  force  equal  to  the  weight  of  the  iron. 

500  g.  x  7.8  =  3,900  g. 

26    (1)  Weight  of  both  in  air     ......  41.2  g. 

(2)  "        "  "     water     .....  26.2  g. 

(3)  "        lost  by  both  in  water      .     .     .  15  g. 

(4)  "  «      iron   "        "    .     .     .     .  _5  g. 

(5)  "  "      cork  "        "    ....  10  g. 

(6)  Density  of  cork  (2.3  -=-  10  =)  .     .     .     .       0.23 


IN  AVERY'S   SCHOOL   PHYSICS.  49 

27.    (a)  See  §  156. 

11.35  =      60°      =  547.13+. 
600  -  w' 

The  lead  weighs  547. 13  +.gr.  in  water. 
(6)  (1)  Weight  of  both  in  air     ....     900  gr. 

(2)  "  "     water      .     .     .     472.5  gr. 

(3)  "         lost  by  both  in  water  .     .     427.5  gr. 

(4)  "  "       lead       "          .     .       52.87  gr. 

(5)  "  "       wood     "          .     .     374.63  gr. 

(6)  Density  of  wood  (300  gr.  -5-  374.63  gr.  =  0.8  +  . 


28.  618  gr. 

31  gr. 


618  gr. 
93  gr. 


649  gr.   -5-   711  gr.  =  0.9-K 

29.  (a)  330  g.  -  315  g.  =  15  g. 

330  g.  -  15  g.  =  22. 

(b)  330  g.  -  303  g.  =  27  g. 
27  g.  +  15  g.  =  1.8. 

(c)  It  displaces  15  g.  or  15  cu.  cm.  of  water. 
Its  volume  is  15  cu.  cm. 

30.  Its  volume  must  be  at  least  that  of  1  Kg.  of  water.    The 
volume  of  1  Kg.  of  water  is  1  liter,  1  cu.  dm.,  or  1,000  cu.  cm. 

31.  See  §160.          2.8  = 


37  —  w' 
.-.  w'  =  23.785+  g. 
Or,  we  may  say  that  the  body  being  2.8  times  as  heavy  as 

QrT 

water,  an  equal  bulk  of  water  would  weigh  '-  —  g.—  13.2142+  g. 

This  is  what  the  body  would  lose  in  water. 

37  g.  -  13.2142  +  g.  =  23.785+  g. 

Page  186. 

1.  15  Ibs.  x  20  x  144  =  43,200  Ibs. 

2.  TrZ)2  =  3.1416  x  16  =  50.2656,    the    number    of    square 
inches  of  surface.     15  Ibs.  x  50.2656  =  753.984  Ibs. 


50  KEY  TO   THE   PROBLEMS 

3.  The  room  contains  6,000  cu.  ft.  or  10,368,000  cu.  in.  of 
air.      This  weighs  0.31    gr.  x  10,368,000  =  3,214,080  gr.,  or 
459.154  Ibs.  Avoirdupois. 

4.  If  the  barometer-tube  had  a  sectional  area  of  1  sq.  cm., 
the  atmospheric  pressure  per  sq.  cm.  would  support  a  mercury 
column  containing  76  cu.  cm.     Such  a  column  of  water  would 
weigh  76  g. ;  such  a  column  of  mercury  would  weigh  76  g.  x 
13.6  =  1,033.6  g. 

Each  side  of  the  cube  has  a  surface  of  100  sq.  cm.  The  six 
sides  have  a  surface  of  600  sq.  cm.  The  atmospheric  pressure 
being  1.0336  Kg.  per  sq.  cm.,  the  total  pressure  is  1.0336  Kg. 
x  600  =  620.16  Kg. 

5.  It  loses  the  weight  of  a  cubic  foot  of  air,  or  about  an 
ounce  and  a  quarter.     §  164. 

6.  (a)  The  trunk  has  an  horizontal  section  of  (2|  x  3J  =) 
8f  sq.  ft.,  or  1,260  sq.  in.      15  Ibs.  x  1,260  =  18,900  Ibs.,  the 
downward  pressure  on  the  top  of  the  trunk.     If  the  trunk  has 
&flat  top,  its  area  will  be  the  same  as  that  of  the  horizontal 
section  (or  bottom)  of  the  trunk ;  if  it  has  an  arched  top,  the 
total  pressure  on  the  upper  surface  will  be  more  than  here 
given,  the  excess  being  lateral  pressure,  which  would  not  at  all 
interfere  with  opening  the  trunk,  even  if  the  air  was  exhausted 
from  it.     The  downward  pressure  would  not  be  affected  by  the 
shape  of  the  top. 

(b)  The  upward  pressure  on  the  under  surface  of  the  trunk 
top  is  equal  to  the  downward  pressure  on  the  upper  surface. 

7.  (a)  None. 

(6)  See  solution  to  Exercise  4. 

8.  (a)  The  capacity  of  the  room  is  320  cu.  m.,  or  320,000  1. 
See  §  164.     1.3  g.  x  320,000  -  416,000  g.,  or  416  Kg. 

(6)  and  (c)  The  surface  is  80  sq.  m.,  or  800,000  sq.  cm 
The  atmospheric  pressure  being  about  1  Kg.  to  the  sq.cm., 
the  pressure  on  each  of  these  surfaces  is  about  800,000  Kg. 


IN  AVERY'S   SCHOOL  PHYSICS.  51 

(d)  The  surface  is  32  sq.  m.,  or  320,000  sq.  cm.    The  pressure 
on  each  of  these  surfaces  is  320,000  Kg. 

(e)  The  surface  is  40  sq.  m.,  or  400,000  sq.  cm.    The  pressure 
on  each  of  these  surfaces  is  400,000  Kg. 

(/)  The  total  surface  of  the  room  is  3,040,000  sq.  cm.  The 
total  pressure  is  3,040,000  Kg. 

(g)  The  outward  pressure  from  within,  is  counterbalanced 
by  the  inward  pressure  from  without. 

9.  A  liter  of  hydrogen  weighs  0.0896  g. ;  10  liters  weigh 
0.896  g.  The  balloon  and  hydrogen  weigh  5.896  g.  The  10  1. 
of  displaced  air  weigh  12.93  g. 

12.93  g.  -  5.896  g.  =  7.034  g. 


Page  19O. 

1.  (a)  Under  a  pressure  of  two  atmospheres. 
(b)    Under  a  pressure  of  half  an  atmosphere. 

2.  Under  ordinary  circumstances,  that  quantity  of  air  would 
weigh  l-i-  ounces.     §  164.     To  get  twice  that  quantity  of  air 
into  that  space,  the  pressure  upon  it  must  be  doubled,  i.e.,  the 
air  must  be  subjected  to  a  pressure  of  two  atmospheres. 

3.  (a)  500  cu.  cm.,  or  half  a  liter. 

(b)  1  liter.     It  makes  no  difference  what  gas  is  used. 

4.  Half  of  it. 

5.  (a)  15  Ibs.  +  5  Ibs.  =  20  Ibs.     §  168. 

(b)  The  pressure  is  30  Ibs.,  or  2  atmospheres.  The 
pressure  being  doubled,  the  volume  will  be  halved,  i.e.,  12 
cu.  in. 

6.  (a)   In  the   short  arm.      The   rising   of   the    barometer 
indicates  an  increase  of  atmospheric  pressure.     This  increase 
of  pressure  will  push  down  the  mercury  in  the  open  arm,  and, 
consequently,  push  it  up  in  the  closed  arm. 


52  KEY   TO   THE   PROBLEMS 

Page    196. 

1.  30  in.  x  13.6  =  408  in.  =  34  ft. 

2.  28  ft. -*- 0.8  =  35  ft. 

3.  76  cm.  x  13.6  =  1,033.6  cm.  =  10.336  m.,  the  height  to 
which  atmospheric  pressure  will  lift  water. 

(a)  and  (b).  It  cannot,  because  atmospheric  pressure  is  not 
sufficient  to  force  the  water  up  to  that  height. 

(c)  It  can,  because  the  water  is  lifted  by  muscular  or  some 
similar  form  of  energy  not  subject  to  the  limitations  placed 
upon  atmospheric  pressure. 

4.  At  the  bottom,  because  the  atmospheric  pressure  being 
greater,  the  resultant  of  the  resultant  forces  (see  §  171)  is 
greater  there. 

5.  34  ft.  -r- 1.8  =  18.8  ft. 

6.  The  stone  is  pushed  up  by  the  pressure  of  the  air  on 
the  under  side  of  the  stone,  when  that  pressure  is  not  counter- 
balanced by  the  weight  of  the  stone  and  the  downward  pressure 
of  the  air  on  the  upper  side  of  the  stone.     That  downward 
pressure  of  the  air  is  largely  carried  by  the  string  that  carries 
the  sucker. 


(6)  There  being  only  -|||-  as  much  air  in  the  receiver 
as  there  was  at  the  beginning,  when  its  elastic  force  was  the 
same  as  that  of  the  external  air,  its  density  and,  hence,  its 
elastic  force  are  only  fff  as  great  as  those  of  the  exter- 
nal air. 

8.  29.5  in.  x  13.6  -f-  1.35  -  223.11  in. 

9.  69  cm.  x  13.6  =  938.4  cm.  =  9.384  m. 

10.    (a)   15  Ibs.  x  3.1416  x  22  =  188.496  Ibs. 
(6)     1  Kg.  x  3.1416  x  42  =  50.2656  Kg. 


IN   AVERY'S   SCHOOL   PHYSICS.  53 

11.  By  placing  the  pump  within  28  ft.  of  the  surface  of  the 
water  and  extending  the  spout  (Fig.  140)  to  the  top  of  the 
well.     The  cylinder  of  the  pump  may  be  a  tube  extending 
to  the  top  of  the  well,  the  piston  rod  running  down  the  inside 
of  such  long  cylinder. 

12.  The  elastic  force  of  the  air,  Avheii  you  blow  in  at  /,  lifts 
the  water  to  li,  and  fills  the  tube.     The  tube  then  constitutes 
a  siphon,  and  continues  to  deliver  water  into  g.     The  rising  of 
the  water  in  g  reduces  the  air  space,  and  thus  increases  the 
elastic  force  of  the  air  in  g,  i  and  a.     This  increased  pressure 
exerted  by  the  air  on  the  surface  of  the  water  in  a  is  trans- 
mitted to  the  water  contained  in  a,  and  forces  it  out  in  a  jet 
at  n. 


54  KEY  TO   THE  PROBLEMS 


CHAPTER   III. 

Page    213. 

1.  See  §  183. 

2.  1,145  -f-  458  =  2.5,  the  number  of  feet. 

3.  The  water  is  a  much  better  conductor  of  sound  than  the 
air  is.     See  §  179. 

4.  The  sound  first  heard  is  propagated  through  the  iron; 
the  other,  through  the  air  in  the  pipe.     The  iron  transmits  the 
sound  waves  more  rapidly  than  the  air. 

5.  The  first  refers  to  the  vibrations  of  the  sounding  body, 
or  to  vibrations  caused  by  them  in  the  air  or  other  medium 
that  may  carry  them  to  the  ear ;  the  second  refers  to  the  sensa- 
tions produced  by  said  waves  through  the  ear  or  organ  of 
hearing. 

6.  See  §  180  (a). 

7.  In  a  vibratory  motion,  each  moving  particle  has  a  to- 
and-fro  motion  so  that  it  periodically  returns  to  the  starting 
point;   translatory  motion  is  progressive  so  that  the  moving 
particle  does  not  return  to  the  starting  point.     The  wind  may 
swing  a  thistlehead  to  and  fro,  giving  it  a  vibratory  motion; 
it  may  also  loosen  some  of  the  winged  seeds,  and  carry  them 
to  another  field  with  a  translatory  motion. 

8.  Because  the  velocity  of  the  particle  changes  like  that 
of  a  common  pendulum.     See  Experiment,  §  108. 

9.  Nowise;  they  are  the  same. 
10.   Sinusoidal. 


IN  AVERY'S   SCHOOL   PHYSICS.  55 

Page  222. 

1.  1,090  ft.  x  18  =  19,620  ft. 

2.  (1,090  ft.  +  2  ft.  x  15)  x  7  =  7,840  ft.     See  §  184  (c). 

3.  Exercise  2  shows  that  the  velocity  of  sound  at  the  given 
temperature   is   1,120   ft.   per  second.     1,120  ft.  -f-  224  =  5  ft. 
See  §  184  (6). 

1,150  - 1,090 

4.  -      — ^ —     —  =60)   the   number   01   centigrade  degrees 

above  the  freezing-point,  i.e.,  30°  C. 

5.  It  acts  as  a  reflecting  surface  to  the  ear,  and  cuts  off 
confusing  sound  waves  from  sources  other  than  the  speaker. 

6.  The  velocity  is  1,120  ft.  per  second.     The  sound  reached 
the  cliff  in  3  seconds.     1,120  ft.  x  3  =  3,360  ft. 

7.  The  velocity  must  be  1,100  ft.  per  second. 

1,100-1,090 

— — -      —  =  5,  the  number  of  centigrade  degrees  above 

z 

the  freezing-point,  i.e.,  5°  C. 

8.  1,120  ft.  -f-  4  ft.  =  280,  the  number  of  vibrations.    The 
given  velocity  implies  a  temperature  of  15°  C.    See  solutions  2 
and  3  above. 

9.  332  m.  =  33,200  cm.     33,200  cm.  +  830  =  40  cm. 

10.  1,128  ft.  -s- 12  ft.  =  94,  the   number   of  vibrations   per 
second. 

11.  It  took  the  sound  T3^  of  a  second  to  reach  the  cliff.     The 
velocity   of    sound   at  the   ordinary  temperature   of  the  air 
(15°  C)  is  1,120  ft.     T3g-  of  1,120  ft.  is  210  ft. 

12.  It  is  a  case  of  echo.     It  took  the  sound  1^  seconds  to 
reach  the  reflecting  surface.     1,090  ft.  x  1 J  =  1,362J  ft. 


56  KEY   TO   THE   PROBLEMS 

13.    Let  x  represent  the  depth  of  the  mine.     From  the  third 

formula   in  §  107,  we    derive    t2  —  —  ,    and   t  =  \/—  .     Hence, 

2x~  9  x  9 

:  represents  the  time  of  falling,  represents   the 

u  1,1^U 

time  required  for  the  sound  to  pass  from  the  bottom  to  the  top 
of  the  shaft.    Adding  the  times  required  for  these  two  motions, 

we  have 

2x  x     _  ~ 

~ 


32.16      1,120 

The  algebraic  solution  of  this  equation  for  the  value  of  x,  the 
required  depth  of  the  mine,  is  somewhat  tedious,  and  may  well 
be  waived  by  the  teacher,  the  mere  statement  of  the  problem  as 
above  being  accepted  as  an  answer. 

14.  See  §  184  (d).     Compare  Exercise  4,  page  214. 

15.  The  prison  was  a  whispering  gallery.     See  §  185. 

16.  The  usual  temperature  of  15°  C.,  and  the  corresponding 
velocity  of  1,120  ft.  per  second,  are  assumed.     The  strokes  fol- 
low at  intervals  of  1  of  a  second  ;  i.e.,  the  pulse  caused  by  any 
stroke  is  i  of  1,120  ft.,  or  224  ft.  in  advance  of  the  pulse  caused 
by  the  succeeding  stroke.     When  the  bells  are  at  the  same 
distance  from  the  hearer,  the  pulses  arrive  at  the  listening  ear 
at  the  same  time,   and  blend  as  a  single  pulse  caused  by  a 
single  stroke.     When  one  bell  is  224  ft.  further  away  than  the 
other,  the  first  pulse  from  the  further  bell  arrives  at  the  same 
time  as  the  second  pulse  from  the  nearer  bell,  and  blends  with 
it   as    a   single  pulse  caused  by  a  single  stroke;  succeeding 
pulses  from  the  two   bells    similarly  blend.     But   when   the 
distances  of  the  bells  differ  by  112  ft.,  the  first  pulse  from  the 
further  bell  arrives  y1^-  of  a  second  after  the  first  pulse  from 
the  nearer  bell  ;  yL  of  a  second  later,  the  second  pulse  from  the 
nearer  bell  arrives  ;  TV  °^  a  second  later,  the  second  pulse  from 
the  further  bell  arrives,  etc.,  in  succession,  the  pulses  arriving 
at  intervals  of  yL  of  a  second  instead  of  blending  in  pairs  as 
before.     Pulses  that  arrive  at  intervals  of  y1^  of  a  second  give 
the  impression  of  10  strokes  per  second. 


IN  AVERY'S  SCHOOL  PHYSICS.  57 

Page  242. 

1.  See  §  194. 

144  x  £  =  172f ,  the  number  of  vibrations  for  the  minor  third. 
144  x  -f  =  180,  the  number  of  vibrations  for  the  major  third. 
144  x  -|  =  216,  the  number  of  vibrations  for  the  fifth. 
144  x  2  =  288,  the  number  of  vibrations  for  the  octave. 

2.  The  octave  above  is  produced  by  (264  x  2  =)  528  vibra- 
tions.    The  fifth  above  the  octave  is  produced  by  (528  X  f  =) 
792  vibrations. 

3.  The  disk  revolves  211  times  per  second.     The  disk  has 
24  holes  in  the  inner  row.     The  number  of  puffs  emitted  per 
second  is  24  x  21^  —  512.     The  unison  indicates  that  the  given 
tone  is  due  to  512  vibrations  per  second.     This  is  twice  the 
number  of  vibrations  assigned  to  middle  C,  so  that  the  tone  is 
just  one  octave  above  middle  C ;  i.e.,  it  is  C4.     See  §  196. 

4.  Evidently  the  vibrations  of  the  two  forks  are  as  9 : 15. 
Referring  to  the  relative  vibration-numbers  as  given  in  §  196, 
we  see  that  27  corresponds  to  the  given  lower  tone  D.     The 
proportion  9  : 15  : :  27  : 45  shows  that  the  higher  tone  is  B. 

5.  Multiply  261  successively  by  the  vibration-ratios  given 
in  §  196. 

6.  The  whistle  has  a  certain  vibration-number  correspond- 
ing to  its  pitch.     When  the  locomotive  and  the  listener  are 
standing  still,  the  number  of  pulses  that  come  to  the  ear  is 
the  same  as  the  vibration-number.     When  the  locomotive  is 
approaching  the  observer,  the  number  of  pulses  per  second 
that  arrive  at  the  ear  exceeds  the  vibration-number,  and  the 
pitch  is  therefore  raised.      When  the  locomotive  is  receding 
from  the  observer,  the  number  of  pulses  per  second  that  arrive 
at  the  ear  is  less  than  the  vibration-number,  and  the  pitch  is 
therefore  lowered.     The  change  of  pitch  is  very  noticeable  if 
the  whistle  is  blowing  when  the  locomotive  passes  by  the 
observer. 


58  KEY  TO   THE   PROBLEMS 

7.  See  the  relative  vibration-numbers  in  §  196. 

36  :  40  :  :  x  :  440 ;  x  =  396,  the  number  of  vibrations. 

8.  Because  the  resistance  of  the  board  reduces  the  velocity 
of  the  saw,  and  the  rapidity  of  the  successive  blows  of  the 
saw-teeth. 

9.  The  rapidity  with  which  the  pulses  would  strike  his  ear 
would  be  doubled,  and  the  pitch  would  be  raisad  an  octave. 

10.  The  rapidity  with  which  the  pulses  would  strike  his  ear 
would  be  lessened.      The  tone  would  be  flatted.     We  might 
imagine  the  velocities  to  differ  so  little  that  the  number  of 
pulses  that  overtook  the  ear  per  second  would  be  so  small  that 
they  would  be  recognized  as  separate  sounds  instead  of  blend- 
ing into  a  continuous  tone. 

11.  The  observer  would  overtake  in  inverse  order  the  pulses 
that  started  ahead  of  him ;  the  music  would  bo  heard  again  as 
if  played  backward. 

12.  The  reed  has  a  certain  vibration-number.     When  the 
observer  is  in  the  prolongation  of  the  axis  of  rotation,  his  dis- 
tance from  the  reed  is  constant,  and  the  number  of  pulses  that 
come  to  his  ear  per  second  is  the  same  as  the  vibration-number 
of  the  reed.     As  the  number  of  pulses  does  nol.  vary,  the  pitch 
is  constant.     When  the  observer  stands  in  th-3  plane  of  rota- 
tion, the  reed  approaches  him,  and  recedes  from  him  once  for 
every  revolution  of  the  axis.     When  the  reed  is  approaching 
the  ear,  the  succession  of  pulses  is  more  rapid,  and  the  pitch 
rises.     When  the  reed  is  moving  away  from  1he  ear,  the  suc- 
cession of  pulses  is  less  rapid,  and  the  pitch  falls. 

Page   254. 

1.  See  §  202. 

2.  Three  beats  per  second. 

3.  The  strings  are  not  in  unison,  or  there  would  be  no 
beats.     The  vibration-number  of  the  second  string  is  398  or 
402  per  second. 


IN  AVERY'S   SCHOOL   PHYSICS.  59 

4.  308-297  =  11. 

5.  See  §  205  (a).     As  the  prong  swings  outward,  it  sends 
a  condensation  down  the  jar.     If  the  air  column  in  the  jar 
is  ^  of  a  wave-length,  the  condensation  will  travel  to  the  lower 
end  of  the  air  column  and  back  in  f,  or  |,  the  period  of  the 
fork.     At  the  end  of  this  half-period,  the  prong  is  beginning 
its  swing  in  the  opposite  direction,  so  that  the  reflected  pulse 
will  coincide  with  the  direct  outward  pulse.     If  the  length  of 
the  air  column  is  f  of  a  wave-length,  the  reflected  pulse  will 
coincide  with  the  direct  pulse  of  the  prong  on  its  second  return 
swing;    if  it  is  j  of  a  wave-length,  the  reflected  pulse  will 
coincide  with  the  direct  pulse  of  the  prong  on  its  third  return 
swing;  etc. 

6.  See  §  205  (d). 

7.  (a)  15  in.  x  4  =  60  in.,  or  5  ft. 

(6)  Assume  that  the  temperature  is  such  that  the  veloc- 
ity of  sound  in  air  is  1,120  ft.  1,120  -j-  5  =  224,  the  number 
of  pulses  per  second.  As  the  vibration-number  is  224,  the 
wave-period  is  2-0-4  of  a  second. 

8.  The  velocity  of  the  sound  is  34,160  cm.     See  §  184  (c). 
The  wave-length  is  64.8  cm.  x  4  =  259.2  cm.     34,160  -5-259.2  = 
131.8,  the  computed  vibration-number.     Under  the  conditions 
assumed  in  the  text,  we  must  allow  for  experimental  error, 
and  take  the  nearest  whole  number  for  the  actual  vibration- 
number. 

9.  Knowing  the  velocity  of  sound  per  second  at  a  given 
temperature,  and  the  number  of  vibrations  per  second  that 
the   sounding   body  undergoes,  it  is  evident  that  the  wave- 
length  may  be  determined   by  dividing   the  velocity  by  the 
vibration-number;  i.e., 


n 


60  KEY   TO   THE   PROBLEMS 

From  §  205  (6),  we  also  know  that  the  wave-length  equals 
four  times  the  length  of  the  resonant  air  column  increased  by 
^  of  its  diameter;  i.e., 

« =<<+!} 

Equating  these  values  of  w,  we  have 

-V-;    whence  v  =  ±(l  +  $}n. 

4:J     n  V       4/ 

10.  w  =  4f  50  +  -  )  =  204,  the  wave-length  in  centimeters. 

11.  There  are  two   beats  per  second.     512  —  2  =  510,    the 
vibration-number  of  the  loaded  fork. 


Page   266. 

1.  Ignore  the  number  of  vibrations,  and  see  §  210. 

2.  (a)    The  vibration-number  is  50.     See  §  210  (1). 
'    (6)    The  vibration-number  is  200. 

3.  See  relative  vibration-numbers,  §  196,  and  §  210  (1). 

27  :  24  : :  18  :  x;  x  =  16,  the  length  in  inches. 

1 00 

4.  =  =  25,  the  vibration-number. 

2xV4 

5.  See  §§  196  and  210  (2). 

3:4::V4:Vz;   »  =  7|lbs. 

6.  3  ft.  :  5  ft.  : :  256  :  x;   x  =  426f,  the  vibration-number. 

7.  2:1::  Vl6  :  V#;    x  =  4,  the  number  of  pounds. 

8.  (a)   The  wave-length  =  1,120  ft. -5-32=  35  ft.    The  length 
of  the  air-column  is  half  the  wave-length.      See   §   212  (a). 
35  ft.  -5-  2  =  171  ft.,  the  length  of  the  tube. 

(6)   The  wave-length  =  1,120  ft.  -=-  4,480  =  J-  ft.,  or  3  in.    The 
length  of  the  air-column  is  half  the  wave-length,  or  1^  in. 


IN  AVERY'S   SCHOOL   PHYSICS. 


61 


9.  (a)  The  length  of  the  air-column  in  an  open  pipe  is 
half  the  wave-length.  4  ft.  -5-  2  =  2  ft.,  the  length  of  the  open 
pipe. 

(6)  The  length  of  the  air-column  in  a  stopped  pipe  is  a 
fourth  of  the  wave-length.  §  212  (a).  4  ft.  -s-  4  =  1  ft.,  the 
length  of  the  stopped  pipe. 

10.  The  catgut  string  will  vibrate  three  times  as  rapidly. 

V9  =  3.     §  210  (4). 

11.  There  would  be  marked  differences  in  pitch.     The  ve- 
locity of  sound  is  greater  for  hydrogen  than  it  is  for  air,  and 
greater  for  air  than  it  is  for  carbon  dioxide.     See  §  184.     For 
a  given  frequency  of  vibration,  the  greater  the  velocity,  the 
greater  the  wave-length.     The  length  of  the  air-column  has  a 
definite  relation  to  the  wave-length.     See  §  212  (a).     Conse- 
quently, a  change  of  gas  that  increases  the  wave-length  must 
be  accompanied  by  an  increase  in  the  length  of  the  organ- 
pipe,  or  by  a   rise  of   pitch.     As   the  velocity  of   sound   in 
hydrogen  is  almost  four  times  as  great  as  it  is  in  air,  the 
tone  emitted  by  a  pipe  filled  with 

hydrogen  is  nearly  two  octaves 
above  that  produced  by  the  same 
pipe  filled  with  air.  For  a  similar 
reason,  the  tone  yielded  by  the  pipe 
when  filled  with  carbon  dioxide  is 
of  lower  pitch. 


12.  Drive  the  handle  of  a  pocket 
tuning-fork  into  a  hole  bored  in  a 
block  of  wood,  and  incline  the  fork 
as  shown  in  the  accompanying  fig- 
ure. Suspend  a  varnished  cork-ball, 
the  size  of  a  pea,  by  a  silk  fiber  or 

very  fine  thread,  sound  the  fork  and  at  once  bring  the  ball 
against  the  face  of  a  prong.  Hunt  for  a  position  in  which 
the  ball  will  remain  at  rest  touching  the  face  of  the  prong. 


62  KEY  TO   THE  PROBLEMS 

One  or  more  such  places  may  be  found  just  above  the  crotch 
of  the  fork.  These  places  of  rest  are  nodes.  Then  slowly 
raise  the  ball,  keeping  it  in  contact  with  the  fork;  it  will 
begin  to  tremble.  Raise  it  higher  until,  near  the  end  of  the 
fork,  it  is  driven  violently  away. 


IN   AVERY'S   SCHOOL  PHYSICS.  63 


CHAPTER   IV. 
Page  274. 

1.  36  -r-  1.8  =  20,  the  difference  in  centigrade  degrees. 

2.  35  x  1.8  =  63,  the  difference  in  Fahrenheit  degrees. 

3.  (a)   68°-32°=36°;  36° -=-1.8=20,  the  centigrade  reading. 

(6)   68°  F. 

4.  10°  C. 

5.  The   mercury  would    stand   higher,    and   the   readings 
would   indicate   temperatures  higher  than   the  real  tempera- 
tures. 

6.  The  error  o>:  the  readings  would  increase  as  the  tempera- 
ture became  higher. 

7.  15°  x  0.8  =  12°,  the  Eeaumur  reading. 

8.  It  would  indicate  that  the  paint  increased  the  thermal 
emission  of  the  tin. 

9.  The  expansion  of  the  flask  increases  its  capacity  before 
the  contained  liquid  expands. 

10.  If  the  bulb  remained  as  large,  a  diminution  of  the  bore 
would  increase  the  sensitiveness  of  the  instrument. 

11.  Perhaps  a  little  vapor  of  mercury ;  otherwise  it  is  a  high 
vacuum. 

12.  The  flow  from  the  warmer  to  the  cooler  body  will  vary 
with  the  difference  of  the  temperatures. 

13.  There  are  none.     See  §  222. 

14.  See  §  222  (a). 


64  KEY   TO   THE   PROBLEMS 

Page  282. 

1.  Because  the  cloth  is  a  poor  conductor  of  heat. 

2.  Because  the  oil-cloth  is  a  better  conductor  of  heat  than 
the  carpet  is. 

3.  Non-conductor  in  each  case. 

4.  The  confined  air  is  a  good  non-conductor. 

5.  Because  of  the  convection  currents  thus  set  up.     See 
§§  227  and  228. 

6.  See  answer  to  Exercise  4.     The  animal  heat  is  better 
retained  thereby. 

7.  Copper  is  a  better  conductor  than  tin. 

Page  286. 

1.  See  §  230  (a).     The  contraction  of  the  cooling  iron  forces 
the  parts  of  the  wheel  closely  together ;  the  cooled  iron  holds 
them  there. 

2.  Such  walls  generally  bulge  outward.     See  §  230  (a). 

3.  To  provide  for  the  inevitable  expansion  and  contraction 
caused  by  variations  of  temperature. 

4.  Because  the  contraction  of  the  alcohol  in  cold  weather 
exceeds  the  diminution  in  the  capacity  of  the  measure. 

5.  See  §  228. 

6.  The  surface  temperature  is  0°.     The  water  at  the  bottom 
of  the  pond  at  the  same  time  may  be  4°.     See  §  232  (a). 

7.  273°.     See  §  232. 

12.  273 :  273  +  300  : :  900  :  x. 

13.  273  :  273  +  100  :  :  1,000  : 1,366.3 ;   1,366.3-1,000=366.3. 

14.  273  +  50  :  273  +  15  :  : 1 5,000  :  x. 


IN  AVERY'S  SCHOOL  PHYSICS.  65 

15.  185°  F.  =85°C. 

273  +  85  :  273  +  10  : :  98  :  x. 

16.  273  +  10:273  +  18.7)    ,^.x 

590  :  530  j 

x  =  143.5  +  ,  the  number  of  cu.  cm. 

17.  273:  273 +  60::  231:*. 

18.  The  water  contracted,  and  the  level  in  the  tube  fell  as 
the  temperature  fell  to  4°.     As  the  water  cooled  from  4°  to  0°, 
it  expanded,  and  the  level  rose.     See  §  232  (a). 

Page  295. 

1.  By  confining  the  steam,  thus  increasing  the  pressure  on 
the  liquid  and  raising  its  temperature.      See   §  240  (a).     A 
vessel  designed  for  this  purpose  is  called  a  digester. 

2.  See  answer  to  Exercise  1. 

3.  By  pumping  out  the  steam  as  fast  as  it  is  formed,  and 
thus  reducing  the  pressure  on  the  liquid.     See  §  240  (3).     A 
vessel  designed  for  this  purpose  is  called  a  vacuum-pan. 

4.  No. 

5.  It  expands  or  it  would  not  float.     It  resembles  ice  in 
this  respect.     As  it  expands  in  the  mold,  it  fills  every  part  of 
the  mold,  in  consequence  of  which  the  face  of  the  casting  is 
sharply  defined.     Lead  and  iron  when  similarly  cast  lack  the 
clear-cut  outlines  of  ordinary  type. 

6.  They  are  due  to  differences  in  atmospheric  pressure. 
See  §  240  (3). 

7.  See  §  235  (3).    The  melting-point  is  lowered  by  the  pres- 
sure, and. the  ice  melts  at  the  surfaces  of  contact.     The  water 
thus  produced  has  the  temperature  of  the  ice  from  which  it 
came.      When    the    pressure    is    removed,   the    melting-point 
becomes   higher   than  the  temperature  of  the  water,  which, 
consequently,  freezes. 

5 


titi  KEY  TO   THE  PROBLEMS 

8.  The  air  is  said  to  be  too  dry ;  i.e.,  the  quantity  of  watery 
vapor  should  be  increased  as  the  temperature  is  increased. 
This  may  be  done  by  keeping  an  open  vessel  of  water  on  the 
stove  or  in  the  hot-air  chamber  of  the  furnace. 

9.  At  115°.     See  §  235  (1). 

10.  See  §  241. 

11.  See  §  240  (c). 

12.  See  §  240  (6). 

13.  Some  of  the  water  passes  through  the  vessel  and  evapo- 
rates.    The  heat  that  produces  this  change  of  condition  dis- 
appears in  the  process.     See  §  236.     This  withdrawal  of  heat 
lowers  the  temperature  of  the  vessel  and  its  contents.     This 
exercise  and  the  one  following  may  well  be  reviewed  after  the 
study  of  latent  heat. 

14.  When  the  water  freezes  it  gives  out  heat.     See  §  234. 
When  the  ice  melts  the  heat  that  does  the  work  disappears  in 
the  process.     See  §  233. 

'     15.  They  are  cases  of  sublimation.     See  §  240  (d). 
16.   An  increase  of  the  humidity  lowers  the  dew-point. 

Page  3O3. 

1.  Find  the  number  of  calories  that  may  be  furnished  by 
the  several  quantities  of  water  in  cooling  to  any  given  tem- 
perature, as  0°  C. 

1  Kg.  at  40°  gives  40  large  calories. 

2  "  "  30°  "  60  "  " 

3  «  «  20°  "  60  "  " 

4  <c  «  10°  «  40  "  " 
10  "  "  200  "  " 

These  200  heat-units  would  warm  the  10  Kg.  of  water  20° 
above  the  assumed  temperature,  or  to  20°.     It  makes  no  differ- 


IN  AVERY'S   SCHOOL   PHYSICS.  67 

ence  what  temperature  is  chosen  for  the  reduction.  If,  e.g., 
we  try  a  temperature  of  10°  we  shall  have  (30  +-  40  +  30  +  0  =?) 
100  heat-units.  This  quantity  of  heat  will  warm  10  Kg.  of 
water  10°  above  the  chosen  temperature,  or  to  20°.  If  a  still 
higher  temperature  is  chosen  for  the  reduction,  care  being 
given  to  the  algebraic  signs,  the  result  will  still  be  the 
same. 

2.  Instead    of    the    water-gram-degree    unit,    or   calory,   we 
may  use   the  water-pound-degree  unit.     Let  x  represent  the 
specific  heat  of  the  mercury.     The  mercury  was  cooled  from 
20°  to   O.G340,   or   19,366°.      1  x  19.3G6  x  x  =  the    number   of 
heat-units  given  up  by  the  lead ;  1  x  0.634  x  1  =  the  number 
of   heat-units  received  .by  the  water.     19.366  x  =  0.634  ;  x  = 
0.0327. 

3.  Let  x  represent  the  number  of  pounds  of  water  required. 
The  water  will  be  cooled  to  0°  and  will  part  with  85  x  heat- 
units.      The  melting   of    the    ice   will    require   80  x  15  heat- 
units.     85  -x  =  1,200  ;  x  =  14.117  +  . 

4.  The  specific  heat  of  ice  being  0.5,  it  will  require  50  heat- 
units  to  warm    the  ice  to  the  melting-point.     It  will  require 
800  more  to  melt  it.     95  x  =  850 ;  x  =  8.94. 

5.  The   specific   heat   of    steam   is   0.48.      Each   pound   of 
steam,  in  cooling  to  100°,  would  yield  12  heat-units.     These, 
with  the  heat  from  condensation  and  cooling  to  25°,  would 
yield  (12  +  537  +  75=)  624  heat-units.      The    required    quan- 
tity of  steam  must  furnish  624  x  heat-units.     To  warm  the 
ice  to  0°  will   require   20   heat-units.      This,  with   the   heat 
required  for  fusion  and  warming  the  melted  ice  to  25°  C., 
amounts  to  545  heat-units. 

G24aj  =  o45;  a  =  0.87  +  . 

6.  The  gram  of  steam  will  yield  (537  +  100  =)637  calories. 
Each  gram  of  ice  will  require  80  calories.     637  -=-  80  =  7.96, 
the  number  of  grams  of  ice. 


68  KEY  TO   THE   PROBLEMS 

7.  Assume  any  quantity  of  ice,  as  1  gram.    Let  x  represent 
the  initial  temperature  of  the  water.     The  melting  of  the  ice 
requires  80  calories,  which  must  be   supplied  by  1   gram   of 
water  in  cooling  to  0°.      For  1  gram  of  water  to  part  with 
80  calories,  it  must  cool  80°.     Consequently,  the  initial  tem- 
perature must  be  80°  higher  than  its  final  temperature. 

8.  Let  x  represent  the  temperature.      Then,  80  +  x  =  10 
(20 -a?);  a;  =  10.9  +  . 

9.  The  water  can  furnish  60  heat-units  for  melting  ice. 
This  will  melt  f  of  a  pound  of  ice.     The  result  will  be  J  of  a 
pound  of  ice  and  5|  Ibs.  of  ice-cold  water. 

10.  The  steam  can  furnish  6,370  heat-units  in  cooling  to  0°. 
This  heat  must  warm  1,010  g.  of  water,  and  can  raise  it  to 
(6,370  -r-  1,010  =)  6.3°+.     Or,  we  may  let  x  represent  the  tem- 
perature.    Then  10  (637  -  x)=  1,000  x;  x  =  6.3°  +  . 

11.  Let  x  represent  the  resulting  temperature.     537  4-100  — 
x  =  10  x.     537  +-  (100  -  x)  =  10  x;  x  =  58  nearly. 

12.  Iron.  Water. 


Specific  heat,  0.1138 

Weights,  200. 

Changes  of  temperature,  300  —  x 


1. 
1,000. 


x 


6,828  -  22.76  x  =  1,000  x;  x  =  6.67  +  . 

13.  Let  x  represent  the  specific  heat.     The  temperature  of 
the  80  grams  falls  80°. 

6,400  x  =  200  x  (20  -  10)  ;  x  =  0.31 +  . 

14.  The  ice  requires  8,000  heat-units.    Each  pound  of  steam 
will  supply  637  heat-units.     8,000  -j-  637  =  12.55. 

15.  It  will  require   (80  x  5=)  400  water-ouiice-degree  heat- 
units  to  melt  the  snow.     The  water  can  furnish  460  such  units. 
Hence  the  snow  will  all  be  melted,  and  the  water  warmed. 
Let  x  represent  the  resultant  temperature  of  the  water. 

5  (80  -f  x)  =  23  (20  -  x) ;  x  =  2.14°+. 


IN   AVEKY'S   SCHOOL   PHYSICS.  69 

16.  The  steam  must  supply  5,000  water-pound-degree  heat- 
units.     Each  pound  of  steam  can  supply  (537  -j-  90  =)  627  such 
units.     5,000  -j-  627  =  7.97. 

17.  Let    x    represent   the    number   of   grams   of   mercury. 
From  Exercise  2,  it  appears  that  the  specific  heat  of  mer- 
cury is  0.0327. 

150  x  299  x  0.0314  =  0.0327  x;  x  =  43,066.97. 

18.  Let  x  represent  the  resultant  temperature. 

4  [537  +  (100  -  x)-]  =  200  (x  -  10)  -,  x  =  22.29°. 

19.  That  the  specific  heat  of  sulphur  is  0.2. 

20.  The  specific  heat  of  a  body  represents  the  number  of 
calories  required  to  raise  the  temperature  of  1  gram  of  that 
substance  1  degree,  and  will  be  the  same  whatever  the  mass  of 
the  body.     The  thermal  capacity  of  the  body  introduces  mass 
as  a  factor.     See  §  248. 

21.  Much  less,  on  account  of  the  high  specific  and  latent  heat 
of  water.     As  will  appear  further  on  (see  the  answer  to  Exer- 
cise 12,  on  page  389),  the  watery  vapor  in  the  earth's  atmos- 
phere  acts  as  a  trap  for  solar  heat,  or  as  a  huge  terrestrial 
blanket. 

22.  Work  is  done   upon    the   gas;    it   is  liquefied   by   the 
pressure  put  upon  it.     The  latent  heat,  not  being  employed  in 
maintaining  the  aeriform   condition,  appears  as  sensible  heat. 
When  the  liquefied  gas  expands  and  resumes  its  aeriform  con- 
dition, it  does  work  and  the  reverse  thermal  effects  follow. 

23.  See  §  234. 

24.  Work  is  done  upon  the  air,  as  in  Exercise  22,  and  the 
compressed  air  is  heated  thereby.     The  pump  is  heated  by  the 
heated  air.     Exercises  22  and  24  will  be  better  understood  after 
the  section  on  the  relation  between  heat  and  work  has  been 
studied. 


70  KEY  TO   THE  PROBLEMS 

Page  311. 

1.  Because  of  this  great  latent  heat  of  water,  the  processes 
of  melting  ice  and  freezing  water  are  necessarily  slow.  Other- 
wise the  waters  of  our  northern  lakes  might  freeze  to  the  bottom 
in  a  single  night,  while  "the  hut  of  the  Esquimaux  would 
vanish  like  a  house  in.  a  pantomime,"  or  all  the  snows  of  winter 
be  melted  in  a  single  day,  with  inundation  and  destruction. 

3.  The  work  done  by  the  engine  is  (8,540  x  50=)  427,000 
kilogrammeters,  or  427,000,000  grammeters.     As  the  mechan- 
ical equivalent  of  a  calory  is  427  grammeters,  the  work  per- 
formed is  equivalent  to  1,000,000  calories,  or  to  1,000  large 
calories. 

4.  See  §  252.      34,462  -=-  1,000  =  34.462.      The  number  of 
calories  developed  would  raise  the  temperature  of  1,000  g.  of 
water  34.462°. 

5.  80  x  1,390  =  111,200,  the  number  of  feet  (in  vacuo)  that 
the  ice  must  fall. 

6.  See   §    247  (6).     0.1138  x  100  =  11.38,   the    number    of 
water-pound-degree   heat-units   required.     Each   such   unit   is 
equivalent   to   about   1,400   foot-pounds,  a  total   of    (1,400  x 
11.38  =)  15,932  foot-pounds,  or  nearly  enough  to  raise  8  tons 
a  foot  high. 

7.  8,080  x  5  x  1,400  -3-  2,000  =  28,280,  the  number  of  feet. 

8.  88.42%  of  5  Ibs.  =  4.421  Ibs.,  the  quantity  of  carbon. 

5.61%  of  5  Ibs.  =  0.2805  Ibs.,  the  quantity  of  hydrogen. 
8,080  x  4.421  x  1,400  =  50,010,352. 

34,462  x  0.2805  x  1,400  =  13,533,227.4. 
Total  number  of  foot-pounds,  63,543,579.4. 
63,543,579.4  -=-  2,000  =  31,771.8,  the  number  of  feet. 

9.  The  weight  makes  no  difference,  as  an  increase  in  the 
weight  would  increase  the  working  power  and  the  work  to  be 
done  at  the  same  rate.     It  may  be  called  1  gram  (or  anything 
else). 


IN   AVERY'S   SCHOOL   PHYSICS.  71 

That  weight  of  water  heated  100°  would  require  100  calories. 
That  weight  of  lead  heated  100°  would  require  (0.0314  x  100  =) 
3.14  calories,  equivalent  to  1,340.78  grammeters.  See  §  85. 

K.E.=—;  1,340.78  =  :—-;   v  =  162.108,   the  velocity  in 

LI  o  j.y.t) 

meters. 

10.  6(442  -  374)  x  0.056  =  22.848,  the  number  of  water- 
pound-Fahrenheit  degree  heat-units  required  for  heating  the 
tin  to  its  melting-point.  25.6  x  6  =  153.6,  the  number  of 
such  units  required  for  melting  the  tin.  The  total  of  such 
units  required  is  176.448.  The  mechanical  equivalent  of  each 
such  unit  is  778  foot-pounds  (§  251).  778  foot-pounds  x  176.448 
=  137,276.544  foot-pounds. 


72 


KEY   TO   THE  PROBLEMS 


CHAPTER  V. 
Page  323. 

1.  (a)  The  shadow  is  the  frustum  of  a  cone,  the  curved 
surface  of  which  is  bounded  by  rays  that  are  tangent  to  the 
ball.  There  is  no  penumbra,  and  the  crosS-section  increases  as 
the  distance  from  the  ball  increases.  See  Fig.  233. 

(b)  The  rays  that  come  from  each  luminous  point  and  that 
are  tangent  to  the  ball  form  a  limiting  surface  as  above- 
described.  The  umbra  is  the  frustum  of  a  cone,  and  is  sur- 
rounded by  a  penumbra.  See  Fig.  234,  and  the  left-hand  end 
of  the  accompanying  figure.  The  cross-section  increases  with 
the  distance  from  the  ball. 


(c)  The  shadow,  including  umbra  and  penumbra,  is  the  frus- 
tum of  a  cone  as  before.     The  umbra  is  a  cylinder.     Have 
each  pupil  draw  a  figure  resembling  Fig.  234,  replacing  the 
lamp  by  a  ball  at  L,  with  diameter  equal  to  that  of  the  ball,  AB. 

(d)  The  shadow,  including  umbra  and  penumbra,  is  the  frus- 
tum of  a  cone  as  before.     The  umbra  is  a  cone  with  its  base  at 
the  intercepting  ball.     Have  each  pupil  draw  a  figure  as  for 
(c),  but  making  the  ball  at  L  larger  than  the  ball  at  AB,  as  in 
the  right-hand  end  of  the  figure  on  this  page. 


IN   AVERY'S  SCHOOL   PHYSICS.  73 

2.  Water   waves ;    because    the    motion   of    the    particles 
involved  in  the  wave  are  transverse  as  it  is  in  the  case  of 
light,  while  in  sound  they  are  longitudinal. 

3.  See  §  254. 

4.  See  §  265. 

6.  The  diameter  of  the  shadow  will  be  5  times  that  of  the 
coin.     The  area  of  the  shadow  will  be  25  times  that  of  the 
coin. 

7.  The  wall  is  100  times  as  far  from  the  eye  as  the  screen 
is.     The   screen  has  an   area  of   9   sq.  in.     9  sq.  in.  x  1002  = 
90,000  sq.  in. 

8.  See  Experiment  211.     The  distance  of  the  lamp  is  three 
times  that  of  the  candle.    The  lamp  is  (32  =)9  times  as  power- 
ful ;  i.e.,  it  has  9  c.  p. 

9.  See  Experiment  212.     One  is  four  times  as  powerful  as 
the  other. 

10.  The  spot  is  translucent,  and  transmits  more  light  to  the 
eye  than  the  rest  of  the  paper. 

11.  As  the  translucent  spot  transmits  more  light  than  the 
rest  of  the  paper,  it  reflects  less ;  the  same  light  cannot  be 
transmitted  and  reflected. 

12.  There  is  no  penumbra. 

13.  See  answer  to  Exercise  1  (d). 

14.  The  umbra  has  a  finite  length  when  the  luminous  body 
is   larger   than   the  intercepting  body;    under  other  circum- 
stances, its  length  is  infinite. 

15.  (a)   It  will  be  in  the  umbra  near  the  apex. 

(6)    It  will  be  in  the  penumbra  at  one  side  of  the  apex 
of  the  umbra. 

16.  Small  density  and  high  elasticity.     The  formula  given 
in  §  184  applies  to  the  medium  of  any  vibratory  motion. 


74 


KEY  TO   THE  PROBLEMS 


1.  45°. 


Page  34O. 


2.   In    the    accompanying  figure,   the   rays   that   form   the 
image  for  the  right  eye  are  marked  with  single  arrowheads  ; 

,E  those  that  form  the  image  for 
the  left  eye,  with  double  arrow- 
heads. 

3.   (a)  See  §  280  (5) ; 

(b)  See  §  280  (6)  ; 

(c)  See  §  280  (4) ; 

(d)  See  §  280  (2) ; 

(e)  See  §  280  (3). 

6.  See  the  lines  with  double 
arrowheads  in  Fig.  256. 

7.  See  §  275. 

8.  (a)  See  §  283 ;    (b)  see  §  281. 

9.  He  cannot   in   either  case.      Whatever  his  position,  the 
image  will  appear  as  far  back  of  the  mirror  as  the  man  is  in 
front  of  it,  and  the  part  of  the  mirror  used  to  give  a  complete 
image  will  be  half  the  length  of  the  man.     If  the  mirror  is  not 
half  the  length  of  the  man,  it  cannot  give  a  full-length  image 
of  him,  no  matter  what  his  distance  from  it.     Let  MN  repre- 


sent the  mirror,  AB  the  man,  and  ab  his  image.  The  triangles, 
Aei  and  Aab,  are  similar  and  Ae  =  ±Aa;  therefore,  ei  —  \A~B. 
But  ei  is  the  part  of  the  mirror  used  in  forming  the  image. 
Suppose  now  the  man  to  move  either  way,  as  to  CD.  The 
image  appears  at  cd,  and  Ce  —  1  Cc.  Hence,  ei  =  ±cd  or  J  CD. 


IN   AVERY'S   SCHOOL   PHYSICS. 


75 


\v 


Wherever  he  stands,  he  cannot  see  his  complete  image  unless 
the  length  of  the  mirror  is  half  his  own  length  ;  if  it  is  too 
short  in  one  case,  it  will  be  too  short  in  every  such  case. 

10.  60°. 

11.  Let  MN  represent  the  mirror,  EE'  the  two  eyes,  and  ee' 
the  two  images  of  the  eyes.     Of  course,  EE'  are  in  the  same 
horizontal  line.      When  E'  is 

closed,  E  sees  the  image  of  E' 

at  e'.     Placing  the  wafer  at  W, 

hides  e'.     When  E  is  closed, 

E'  would  see  the  image  of  E 

at  e,  were  it  not  for  the  wafer 

which   hides   it.      See    §    276. 

By  drawing  the  parallelogram, 

EE'e'e,  and  remembering  that 

MN  is  midway  between  EE' 

and  ee',  it  may  be  proved,  geometrically,  that  Ee'  and  E'e  will 

intersect  at  W  as  represented  in  the  figure. 

12.  Construct  the  figure,  and  measure  the  lengths  of  object 
and  image.     Modify  Fig.  256,  by  placing  OB  as  far  at  the  left 
of  the  mirror  as  C  is  at  the  right. 

13.  f  =     ' Either  conjugate  focal  distance  equals  the 

product  of  the  other  conjugate  focal  distance  into  twice 
the  principal  focal  distance,  divided  by  the  difference  between 
twice  the  other  focal  distance  and  twice  the  principal  focal 
distance.  Notice  that  r  is  twice  the  principal  focal  distance. 

14.  Since  the  sum  of  the  reciprocals  of  /  and  /'  is  constant, 

2 
i.e.,  -,  it  necessarily  follows  that  if  one  increases   the  other 

must  decrease.  When  either  conjugate  focal  distance,  as  /, 
becomes  infinite,  its  reciprocal  becomes  zero.  Substituting 
this  value  in  the  formula  given  in  §  280  (c),  we  have 

1919  v 

0  +  -  =  -:     -  =  -:     f  =  -- 

V      r'    f      r'   J       2 


76 


KEY  TO   THE  PROBLEMS 


15.  Such  a  negative  value  indicates  that  the  focus  is.  back  of 
the  mirror  ;  i.e.,  that  it  is  virtual  instead  of  real. 

16.  Locate  the  three  points  on  paper,  and  letter  them.    Draw 
the  lines,  AC  and  BC.     Bisect  the  angle,  ACB,  by  the  line, 
CD.     Through  C,  draw  the   line,   mr,   perpendicular  to  CD. 
This  line,  mr,  indicates  the  position  of  the  mirror.     The  angle, 
ACD,  is  the  angle  of  incidence.     The  angle,  BCD,  is  the  angle 
of  reflection. 

17.  If  the  middle  of  a  plane  mirror  is  placed  opposite  the 
corner  of  a  building  so  that  the  line  of  the  face  of  the  mirror 
makes  an  angle  of  45°  with  the  prolongation  of  the  sides  of 
the  house,  as  shown  in  the  accompanying  figure,  a  person  at  A 
might  see  an  image  of  a  person  at  B.     A  rectangular  box  may 


be  made  with  openings,  as  shown  at  a  and  b,  in  the  second 
figure  accompanying.  Mirrors  may  be  placed  so  as  to  make 
angles  of  45°  with  the  sides  of  the  box,  as  shown  at  mmmm. 
An  object  at  a  may  be  seen  by  an  eye  at  b,  the  light  being  re- 
flected by  each  mirror.  The  experiment  is  made  more  interest- 
ing by  placing  an  opaque  object,  as  a  brick,  at  B. 

18.  It  must  be  vertical. 

19.  Two  mirrors  at  an  angle  of  60°  with  each  other. 

20.  If  the  rays  that  are  reflected  by  the  outer  parts  of  the 
mirror  are  removed,  the  foci  of  the  others  will  not  vary  much 
from  F. 

21.  Try  it  with  a  bit  of  looking-glass  and  a  pin.     The  image 
will  be  horizontal. 


IN  AVERY'S   SCHOOL   PHYSICS. 


77 


22.  See  §  270.  The  small  particles  of  water  suspended  in 
the  air  reflect  light,  but  only  those  that  are  in  certain  positions 
relative  to  the  observer  can  reflect  light  to  his  eye.  The  circle 
sometimes  seen  around  the  moon  is  similarly  produced.  Com- 
pare §§303  and  315  (a). 


FIG.  A. 


Page  363. 

1.    In  the  accompanying  figure,  ab  represents  the  plane  of 
the  horizon  for  an  observer  at  E.     A  ray  of  light  coming  to  the 

eye   from    the    sun 

tf  ™.;n    i^    r^^A       *~ s^^^>^ 


at  $  will  be  grad- 
ually bent,  describ- 
ing a  curve.  The 
direction  in  which 
the  sun  is  seen  is 
that  of  a  tangent 
to  this  curve  at  the 
eye.  This  makes  the  sun  appear  higher  than  its  true  position, 
as  at  /S".  Similar  lines  drawn  toward  the  right  from  E  will 
represent  the  morning  phenomenon. 

2.    See  §  290  («).     The  pupil  is   supposed  to  know  that  a 
burning-glass  is  a  single  convex  lens.     If  he  does  not  know  it, 
the  teacher  should  send  him  to  the  dictionary  or  other  source 
i  of  information. 

4.  (a)  See  §  285,  and  Ex- 
periments 233  and  234. 

(b)  See  Tig.  271,  especially 
the  limiting  lines,  Aa  and  Ee. 


5.  See  §§   285  <7>)  and   284 
(c).       The    ratio    of    the    two 
radii  is  4  :  3.     The  angle  BAD 
is  the  angle  sought. 

6.  See  §  288  (6). 


Fia.  B. 


78 


KEY  TO   THE  PROBLEMS 


7.  Draw  CC'  through  the  centers  of  curvature.  Draw  AB, 
the  incident  ray,  and  BC.  The  problem  is  to  construct  BD 
and  DA',  the  paths  of  the  ray  through  the  lens  and  beyond. 
With  B  as  a  center  and  with  radii  of  2  and  3,  draw  the  arcs,  n 
and  m.  From  the  intersection  of  n  and  AB,  draw  a  line 
parallel  with  BC,  and  cutting  the  arc,  m.  From  the  point 

where  this  line 
cuts  m,  draw  a 
line  through  72 
to  D.  The  line, 
BD,  is  the  path  of 
the  ray  through 
the  lens.  Draw 
DC1.  From  D  as 
a  center  and  with 
radii  of  2  and  3, 
draw  the  arcs,  y  and  x.  Produce  BD,  and  from  its  intersection 
with  x,  draw  a  line  parallel  to  DC'  and  cutting  the  arc,  y. 
Through  the  point 
where  this  line  cuts 
y,  draw  DA',  which 
line  marks  the  path 
of  the  ray  beyond 
the  lens. 


-„. 

^  '•f*T''=-fc-'>->P/\p  ---"*"  ''\ 


8.  See   Fig.    D. 
Draw  AB  parallel  to 
CC',  and    complete 
the  construction  as 
in  Exercise  7. 

9.  See    Fig.    E. 
Draw  AB  parallel  to 
the  principal  axis  as 
before,  and  complete 

the  construction.     The  focus  at  F  is  virtual. 


See  §  292,  b  (1). 


AVERY'S   SCHOOL   PHYSICS. 


79 


10.  The  optical  center. 

11.  See  §  291  (6). 

(d)   The   method   of  construction    is    fully    illustrated   by 
Fig.  F.     The  effect  of  moving  the  object  toward  the  principal 


FIG.  F. 


focus   and   the   lens,  the  image   receding   as   the   object    ap- 
proaches, is  illustrated  by  Fig.  Gr. 


FIG.  G. 


(e)   The   emergent   rays   are   parallel,  as  is  illustrated   by 
Fig.  H,  and  form  no  image. 


H. 


80  KEY  TO   THE   PROBLEMS 

(/)  Magnified,  erect  and  virtual,  as  illustrated  by  Fig.  I. 


FIG.  I. 

12    (a)  See  §  291  (6)  (3). 

(6)   See  §  291  (ft)  (2).     Construct  the  image  on  the  scale 
of  1  : 1. 

13.   The  focus  is  virtual.     See  Fig.  J. 


FIG.  J. 

14.  Secondary  foci ;  they  will  be  equal ;  construct  the  image. 

15.  Inasmuch  as  the  incident  rays  are  diverging  and  the 
focus  is  real  instead  of  virtual,  it  is  apparent  that  the  lens  is 
convex.     See  §  292  (7>)  (2).     The  focal  distance  is  8  inches. 

16.  (a)  Five  feet  from  the  flame.     See  §  291  (6)  (4).     Notice 
the  principle  of  reversibility  that  prevails  in  optics. 

(5)  One   will   be   five   times   as   long  as  the   other.     Their 
lengths  are  proportional  to  their  distances  from  the  lens. 

17.  Make  the  experiment,  and  see  Experiment  230. 


IN   AVERY'S   SCHOOL   PHYSICS.  81 

18.  Part  of  the  incident  light  is  reflected  at  each  surface  of 
eat-h  particle.  What  escapes  reflection  by  one  particle  is 
reflected  by  some  other  particle.  All  of  the  light  being  thus 
reflected,  none  is  transmitted,  and  the  glass  is,  therefore, 
opaque. 

Page  389. 

1.  Because  of  the  uniform  angular  distance  from  the  axis 
of  the  bow,  as  explained  in  §  303  (c). 

2.  See  §  299  (a).     0.007  x  0.3937. 

3.  186,000  x  5,280  x  12  -  0.00002. 

4.  See  §§  294  and  298  (a). 

5.  Dispersion.     See  §  368. 

6.  See  §  303  (a). 

7.  See  §§  307  and  308  (3). 

8.  See  §  308. 

9.  Because  the  air  is  diathermanous ;  i.e.,  it  transmits  the 
radiant  energy  instead  of  absorbing  it. 

10.  The  glass  is  diathermanous  to  the  short-wave  radiations 
that  constitute  the  sunshine,  and  athermanous  to  the  long-wave 
radiations  emitted  by  the  objects  within  the  greenhouse  after 
they  have  been  heated  by  absorbing  the  solar  energy. 

11.  Because  the  watery  vapor  in  the  atmosphere  transmits 
to  the  earth  the  short-wave  radiations  of  the  sun  and  absorbs 
the  long-wave  radiations  from  the  earth.     See  §  309(6). 

12.  Because   of    the    diathermancy    of    dry    air    and    the 
athermancy  of  moist  air,  as  stated  in  §  309  (b).     When  the 
night  is  clear,  the  earth's  heat  is  rapidly  radiated  into  space ; 
clouds  act  as  a  blanket  and  absorb  the  energy  radiated  from 
the  earth,  and  return  some  of  it  to  the  earth.     §  310. 

13.  The  glass  is  diathermanous  to  short-wave  radiations.    As 
the  energy  is  freely  transmitted,  it  cannot  also  be  absorbed. 
Only  the  absorbed  energy  heats  the  medium. 


82  KEY  TO  THE  PROBLEMS 

14.  That  water  transmits  short-wave  radiations  and  absorbs 
long-wave  radiations. 

15.  "  Radiant  heat  "  is  used  as  a  synonym  for  radiant  energy, 
and  "  obscure  heat "  as  a  synonym  for  radiations  of  greater  wave- 
length than  any  that  constitute  a  part  of  the  visible  spectrum, 
neither  of  which  is  heat  at  all.     See  §  225. 

16.  If  the  radiations  from  the  lamp  are  passed  through  a 
solution  of  iodine  in  carbon  disulphide,  the  short-wave  energy 
will  be  absorbed  and  the  long-wave  energy  transmitted.     See 
§§  309  (b)  and  313.     If  the  radiations  are  passed  through  a 
solution  of  alum  in  water,  the  long-wave  energy  will  be  absorbed 
and  the  short-wave  energy  transmitted. 

17.  Because  its  polished  surface  is  a  poor  radiator  (§  311). 

18.  The   convection  currents   of  heated  air  have  varying 
densities  and  refractive  powers. 

19.  See  answer  to  Exercises  11  and  12. 

Page  392. 

10.  A  Plticker-tube  is  a  Geissler  tube  (Fig.  409)  with  a  capil- 
lary part,  by  means  of  which  the  luminous  intensity  of  feeble 
electric  discharges  may  be  raised  sufficiently  to  allow  of  spectro- 
scopic  investigation. 


IN   AVERY'S   SCHOOL  PHYSICS. 


CHAPTER  VI. 
Page  43O. 

1.  See  Experiment  298. 

2.  See  §  329  (a). 

3.  See  §  329. 

4.  The  opposite  character  of  the  electrifications  is  indicated 
by  the  attraction  between  the  nibbed  body  and  the  body  with 
which  it  is  nibbed ;  the  equality  of  the  two  electrifications  is 
indicated  by  the  fact  that  when  the  two  bodies  are  brought 
together,  the  two  electrifications  mutually  neutralize  each  other. 

5.  To  avoid  the  condensation  of  moisture  from  the  air.     See 
§  326  (a). 

6.  Since  the  charges  are  of  opposite  signs,  the  force  will  be 
attractive  and  not  repellent.     See  §  332. 

Qxry_24x8_1? 

d2  42 

Our  units  are  all  C.G.S.  units.     By  recognizing  the  algebraic 
signs,  we  have  +24x(-8)_ 

42 

which  also  indicates  an  attractive  force. 

7.  The  —  8  units   will  neutralize  an  equal  number  of  the 
+  units,  leaving  -f- 16  units  to  be  equally  divided  between  the 
two  balls  (which  are  assumed  to  be  of  equal  capacity). 


d2  42 

As  the  algebraic  sign  of  the  answer  is  +  (the  charges  being 
alike),  the  force  is  one  of  repulsion. 


KEY   TO   THE   PROBLEMS 


.'.32  = 


28^<_56 
d2 


Whence,  d2  =  28  x. 56  =  49. 


32 


=  7. 


9.  (a)  See  Fig.  321.  Place  the  dozen  globes  in  actual  con- 
tact. They  will  be  polarized  as  a  single  body,  and  may  all  be 
charged  as  described  in  Experiment  306.  (6)  Charge  one 
negatively  by  induction ;  with  it,  charge  another  positively  by 
induction.  For  example,  give  a  positive  charge  to  (7  and  bring 
it  near  M,  thus  polarizing  the  latter  as  shown  in  the  figure. 


Touch  M  with  the  finger,  thus  leaving  it  negatively  charged. 
Remove  (7,  and  bring  N  into  position,  as  shown  in  the  figure. 
N  will  be  polarized  by  the  negative  charge  of  M.  Touch  N 
with  the  finger  and  its  negative  electrification  will  be  repelled 
to  the  earth,  leaving  the  conductor  positively  charged. 


xo. 


11.  By  bringing  the  charged  conductor  into  simultaneous 
contact  with  two  conductors  of  equal  capacity. 

12.  See  §  333  (a). 

13.  See  §  343  (a). 


IN  AVERY'S  SCHOOL   PHYSICS.  85 

14.  The  shock  will  be  greater  when  the  jar  is  held  in  the 
hand.     When  the  outer  coat  is  insulated,  its  charge  "binds" 
part  of  the  charge  of  the  inner  coat. 

15.  The  divergence  will  be  greater  when  the  jar  is  held  in 
the  hand,  for  the  reason  just  given. 

16.  When  an  electrified  glass  rod  is  brought  near  an  electric 
pendulum,  the  pith-ball  is  polarized,  as  shown  in  the  figure. 
As  the  distance  between  the  opposite  electrifications  is  less 
than  the  distance  between  the  similar  elec- 
trifications, the  attraction  is  greater  than 

the  repulsion.      If   the   pith-ball   is   sus- 
pended, not  by  a  silk  thread  but  by  somo 
good  conductor,  the  attraction  will  be  more 
marked,  for  the  +  of  the  ball  will  escape  to  the  earth  through 
the  support,  and  thus  the  repelling  component  will  be  removed. 

Page  442. 

1.  The  table  gives  it  as  1.023  ohms. 

2.  The  resistance  of  1,000  feet  of  No.  4  copper  wire  is 
0.254  of  an  ohm ;  that  of  800  feet  of  such  wire  is  0.1932  of  an 
ohm.     Multiplying  this  by  the  ratio  between  the  resistivities 
of  copper  and  of  German-silver  as  given  in  the  Appendix  (page 
592),  we  have  0.1932  x  128.29-^-10.45  =  2.374,  the  number  of 
ohms. 

3.  The  resistance  of  that  length  of  No.  8  copper  wire  is 
0.643  of  an  ohm  x  0.750  =  0.48225  of  an  ohm.    Multiply  this  by 

f*  Q  Q  ^ 

— '-—;.  the  ratio  between  the  resistivities  of  copper  and  of  iron. 
10.45 

4.  The  resistance  of  that  length  of  No.  14  copper  wire  is 

O  Q  A 

2.585  ohms  x  0.35  =  0.90475  of  an  ohm.    Multiply  this  by  ~ 
the  ratio  between  the  resistivities  of  copper  and  of  silver. 

5.  33.135  ohms  x  6.050  =  200.46675  ohms. 


86  KEY  TO   THE  PROBLEMS 

6.  The  table  shows  that  386.8  ft.  of  No.  14  copper  wire 
has  a  resistance  of  1  ohm.     The  given  resistance  of  1.75  ohms 
indicates  that  the  fault  would  be  386.8  ft.  x  1.75  =  676.9  ft. 
from  the  end  of  the  line  if  the  wire  was  of  copper.     Correct 

this  result  by  multiplying  676.9  ft.  by  —  :  —  ,  the  ratio  between 

63.35 
the  resistivities  of  iron  and  of  copper. 

7.  The  resistance  of  3,590  ft.  of  No.  14  copper  wire  is  2.585 

ohms  x  3.59  =  9.28  ohms.      Multiply  this  by  63'35,  the  ratio 

10.45 
between  the  resistivities  of  copper  and  of  iron. 

8.  The  length  of  a  No.  14  copper  wire  that  has  a  resistance 
of  1.75  ohms  is  (386.8  x  1.75=)  676.9  ft.,  as  given  in  the  answer 
to  Exercise  6. 

9.  The  table  shows  that  the  resistance  of  No.  13  copper 
wire  is  very  nearly  2  ohms  per  thousand  feet,  and  that  the 
diameter  of  such  a  wire  is  0.071961  of  an  inch,  or  1.828  milli- 
meters.    For  exact  determination,  apply  the  principle  set  forth 
in  §  352  (2)  :     2  ohmg  .  2M8  ohmg  .  .  (L828)2  .  ^ 

The  value  of  x  will  represent  the  diameter  in  millimeters. 
2  ohms  :  2.048  ohms  :  :  (71.961)2  :  y*. 

rj?he  value  of  y  will  represent  the  diameter  in  mils. 

A  shorter  solution  may  be  had  by  using  the  formula  for 
resistivity,  given  on  page  592  : 

2  =  10.45x1,000.   d  =  72+mils. 

The  difference  in  the  two  results  is  partly  due  to  the  fact  that 
the  resistivity  as  given  in  the  table  was  computed  for  the  tem- 
perature of  20°,  while  the  resistance  was  computed  for  the 
temperature  of  24°. 

10.    See  the  formula  for  resistivity  given  on  page  592. 


ou 
This  resistivity  indicates  German-silver  as  the  metal. 


IN   AVERY'S   SCHOOL   PHYSICS.  87 

Page  4-5O. 

5.  See  §  81  (a). 

746  watts  =  1  H.P.  =  33,000  foot-pounds  per  minute. 
33,000  -=-  746  =  44.236. 

6.  9.6  x  2,900  -r-  746  =  37.32. 

7.  37.32  -  0.90  =  41.46. 

8.  (a)  0=^;     21  =  —  ;     72  =  0.07143. 

R  R 

(b)  21  x  1.5  =  31.5. 

(c)  There  will  be  1,600  ft.  of  line  wire,  the  total  resist- 
ance of  which  is  indicated  by  the  answer  to  (a).     This  repre- 
sents a  resistance  per  1,000  ft.  that  is  less  than  that  of  the 
largest  wire  mentioned  in  the  table. 

9.  (a)  110  x  0.5  -r- 16  =  3.4375. 
(6)  45x10-2,000  =  0.225. 

10.  There  is  no  difference  : 

900  x  10  =  1,800  x  5. 

11.  110  x  0.5  x  50  -5-  0.90  -r-  746  =  4.09. 

12.  See  §  361  (6) ;  W=  R  x  C2  =  23  x  3.52  =  281.75. 

13.  TT=— =  —  =  327.02. 

14.  (a)  350,000  =  35  the  number  of  amperes. 

10,000 

(b)  35  x  14  =  490,  the  number  of  volts. 

(c)  490  x  35  =  17,150,  the  number  of  watts. 

(d)  The  525  H.P.  expended  on  the  dynamo  is  equivalent 
to  (746  x  525  =)  391,650  watts. 

350,000  -5-  391,650  =  0.893+ .  Ans.  89.3+  per  cent. 


88  KEY  TO   THE   PROBLEMS 

Page  476. 

1.  The  non-active  part  that  is  about  midway  between  the 
poles.     See  §  367. 

2.  See  §  369  (d). 

3.  By  the  rapidity  of  oscillation  after  the  needle  has  been 
displaced  from  its  chosen  position.     The  greater  the  intensity 
of  the  field,  the  greater  the  rapidity  of  oscillation.     "  When- 
ever a  body  oscillates  under  the  action  of  a  force,  the  square  of 
the  time  of  a  single  oscillation  is  directly  proportional  to  the 
moment  of  the  inertia  of  the  body  about  the  axis  of  oscillation, 
and  inversely  proportional   to  the  directive   force."     As   the 
number  of  oscillations  is  the  reciprocal  of  the  time  or  period 
of  an  oscillation,  it  follows  that  the  directive  force  is  directly 
proportional  to  the  square  of  the  number  of  oscillations.    Com- 
pare the  formula  in  §  115. 

4.  See  Laboratory  Exercise  2,  on  page  476.    No  magnet  can 
be  found  that  tends,  as  a  whole,  to  drift  toward  the  north  or 
south.     The  attraction  of  the  north  pole  of  the  earth  for  the 
marked  pole  of  the  needle  is  counterbalanced  by  its  repulsion 
for  the  unmarked   pole  of   the   needle.      The   accompanying 
figure  shows  the  way  in  which  the  needle  is  placed  in  a  north 

and  south  line.  The  arrow  a 
represents  the  attraction  of 
the  north  magnetic  pole  of  the 
earth,  and  c  represents  the  re- 
pulsion of  the  south  magnetic 
pole  of  the  earth  acting  upon 
the  marked  pole  of  the  needle ; 
b  represents  the  repulsion  of  the  north  pole,  and  d  represents 
the  attraction  of  the  south  pole  of  the  earth  acting  upon  the 
unmarked  pole  of  the  needle.  The  combined  effect  of  these 
four  forces  is  to  place  the  needle  in  the  magnetic  meridian. 
When  the  needle  is  thus  placed,  since  the  two  ends  of  the 
needle  are  practically  equidistant  from  either  pole  of  the 


IX  AVERY'S   SCHOOL   PHYSICS.  89 

earth,  a  +  c  =  b  +  d.     The  two  couples  thus  counterbalance 
each  other. 

5.  See  §376.     220x3-0.7958  =  829+. 

6.  (a)  40  x  5  -T-  0.7958  =  251.3,  the  number  of  gilberts. 

(b)  251.3  -r-  0.00593  =  42,378,  the  number  of  webers. 

(c)  See  §  369  (a).     43,378  -=-  3  =  14,126,  the  number  of 

gausses. 

(d)  See  §  378.     T1L¥  =  0.000593,  the  reluctivity. 

(e)  251.3  -r-  30  =  8.37,  the  number  of  gausses. 

Page  482. 

=  2-+0.1^  = 

o 


'0.5 


+  500^  =  0.004  nearly. 


=  10  -s- (2.5  +  0.1)=  3.846+. 

=  10  -(2.5  +  500)==  0.0199+. 
nr  +  R 

3.  (7=  2 -f- (1  +  16)  =0.1176. 

4.  C=  16-*- (64 +  16)  =0.2. 

5.  0=8-5-  (16  +  16)  =  0.25. 

6.  C=l-f- 5.001  =  0.19996+. 

7.  C=  1-5- (0.5  +  0.001)  =1.996+. 

8.  C  =  10  -r-  (50  +  0.001)  =  0.19999+. 

9.  C  =  1  -«-  (5  +  1,000)  =  0.00099502. 

10.  <7  =  1  -&-  (0.5  +  1,000)  =  0.0009995. 

11.  C  =  10  -=-  (50  +  1,000)  =  0.00952,  the  number  of  amperes. 

NOTE.  — Compare  the  results  in  Exercises  6,  7,  and  8,  where  we  have 
a  small  external  resistance.  Then  compare  the  results  in  Exercises  9, 
10,  and  11,  where  we  have  a  high  external  resistance. 


90  KEY   TO   THE   PROBLEMS 

12.  (a)  Consult  the  table  in  the  Appendix.     0.643  x  0.006 
=  0.00386 ;  0.8  x  6  +  0.00386  =  4.80386. 

(b)  C  =  12  --  4.80386  =  2.498. 

13.  (a)  03  +  0.00386  =  0.13719. 

6 

(6)    0=2-5-  0.13719  =  14.6  nearly. 

14.  The  pupil  must  judge  from  the  evident  analogy,  "tan- 
dem" referring  to  grouping  in  series,  and  "abreast"  referring 
to  grouping  in  parallel. 

15.  Test  them  with  a  high  resistance. 

16.  Apply  the  "  rule  of  thumb  "  given  in  Experiment  347, 
and  the  "  cork-screw  rule  "  given  on  page  468  to  determine  the 
direction  of  the  lines  of  force.     Consider  the  rectangle  as  a 
disk  magnet  (see  Experiment  348),  and  compare  its  action  with 
that  of  a  magnetic  needle.     Consider   the   solenoid  as  a  bar 
magnet,  determine  its  polarity,  and  see  §  371. 

Page  496. 

1.  See  §  388. 

2.  See  §  395. 

3.  See  Experiments  315  and  347. 

4.  See  Experiments  353  and  357. 

5.  The  cast  iron  has  a  residual  magnetism  that  is  helpful 
in  starting  the  machine  (§  396  a),  and  it  is  cheaper  and  more 
easily  worked  than  wrought  iron  or  pure  soft  iron  is.     In  most 
other  respects,  a  pure  soft  iron  core  would  be  vastly  preferable. 

6.  It  may  depend  upon  the  intensity  of  the  magnetic  field, 
the  number  of  coils  on  the  armature,  or  the  rapidity  of  revolu- 
tion of  the  armature. 

7.  The  field  magnets  of  one  are  permanent  5  those  of  the 
other  are  temporary. 

a   It  diminishes  it,  by  increasing  the  resistance  of  the  wire. 
9.  See  Fig.  388, 


IN  AVERY'S   SCHOOL   PHYSICS.  91 

Page  518. 

1.  (a)  See  §  401.     (6)  and  (c)  See  §  400  (a). 

2.  See  §  407. 

3.  See  §  400.     It  weakens  the  direct  current  so  that  it  may 
be  unable  to  force  its  way  through  the  dielectric,  as  in  the  case 
of  the  sparks  from  an  induction  coil. 

4.  (a)  See  §§  378  and  379;  oersteds  =  gl]kerts. 

webers 

(6)  Draw  a  right-angled  triangle;  mark  the  hypotenuse 
"  impedance,"  and  the  other  two  sides  "  reactance  "  and  "  re- 
sistance." See  §  401  (a). 

5.  See  Fig.  399  and  §  403(6). 

6.  See  §  404. 

Page  536. 

1.  See  Fig.  425.     Represent  the  current  that  flows  through 
nx  by  the  letter  c;  and  the  current  that  flows  through  mp  by 
c'.     See  §  361 ;  E=Cx  R. 

The  fall  of  potential  from  A  to  C  =  en;  represent  it  by  v. 
The  fall  of  potential  from  A  to  D  =  c'm;  represent  it  by  v'. 
But  when  the  resistances  are  adjusted  so  that  there  is  no 
flow  through  G,  C  and  D  are  at  the  same  potential ;  i.e.,  v  =  v'. 

Hence,  en  =  c'm,  and  —  =  —  • 
m      c 

x      c' 
Similarly,  it  may  be  proved  that  -  =  -•      Equating  these 

i  P      c 

values  of  -,  we  have  —  —  x,  or  m  :  n  : :  p  :  x.  Q.E.D. 

G  m     p 

2.  That  they  may  absorb  only  a  small  part  of  the  energy. 
If  either  instrument  absorbed  much,  it  would  change  the  value 
of  the  function  that  it  is  to  measure.     A  voltmeter  receives 
the  full  voltage  ;  it  should  use  a  small  current ;  i.e.,  the  resist- 
ance should  be  high.     An  ammeter  receives  the  full  current ; 
its  resistance    should  be  low  so  that  the  voltage  required  to 
force  the  current  through  it  may  not  be  excessive. 


92  KEY  TO   THE   PROBLEMS 


3.   (a)  C  =  =-  =  n/>  '      =  0.001423,  the  number  of  amperes. 
R      26,000 

(6)  E=CR  =  0.003  x  26,000  =  78,  the  number  of  volts. 


amperes. 

(6)  40,000  :  25,000  :  :  110  :  68.75,  the  number  of  volts. 
(c)   40,000  :  15,000  :  :  110  :  41.25,  the  number  of  volts. 


R     3,500  +  2 
peres. 

(b)  E=CR  =  0.004  x  3,500  =  14,  the  number  of  volts. 

(c)  110-14  =  7.85. 

(d)  117  -r-  27,500  =  0.004254,  the  number  of  amperes. 

(e)  The  voltmeter  indicates  (.#=  (7/2=0.004254x117=) 
14.89  volts.     117  -r-  14.89  =  7.85,  the  same  multiplier  as  before. 

6.  (10  -  1)  X  3,500  =  31,500,  the  number  of  ohms. 

7.  110  x  0.002  =  0.22,  the  number  of  watts. 

8.  6.435  :  0.065  : :  6  :  x;  x  =  0.0606  of  an  ohm. 

9.  Pass  a  current  of  known  strength  through  the  coil  to  be 
measured.     Take  the  readings  of  a  voltmeter  at  the  terminals 
of  the  coil.      Substitute  the  known  values  in   the   formula, 

B  =  —,  and  solve  for  the  value  of  R. 
G 

10.   See  Fig.  387.     Connect  the  terminals  of  the  voltmeter 
to  the  main-circuit  wire  each  side  of  the  field  magnets. 

Page  575. 

1.  C=  —  = 838'44 =  10.04.   The  problem  ignores 

R     4.56  x  16  +  10.55 

the  resistance  of  the  line,  i.e.,  assumes  that  the  circuit  is  short 
and  of  inconsiderable  resistance. 

2.  See  §  400. 


IN  AVERY'S  SCHOOL  PHYSICS.  93 

3.  See  §414(». 

9.925  :  x  :  :  tan  60°  :  tan  74°. 
9.925  :  x  :  :  1.73  :  3.49  ;  x  =  20+. 

4.  R  =  —  =  —  =  30,  the  number  of  ohms. 

C      1 


5.   (a)  a  =     =        =  0.46  nearly. 


(&)   110  x  0.46  =  50.6. 
(c)   50.6  -=-  16  =  3.16. 

6.  The  resistance  of  the  series  of  lamps  will  be  250  ohms. 
That  of  the  wire  may,  then,  be  5  ohms.     This  is  at  the  rate  of 
25  ohms  per  1,000  ft.    No.  24  is  the  nearest  to  the  size  desired, 
but  as  the  line  resistance  "  must  not  be  more  "  than  5  ohms, 
we  must  use  the  next  larger  size,  or  No.  23,  copper  wire. 

7.  The  resistance  of  the  lamp  circuit  will  be  2.5  ohms,  and 
that  of  the  200  ft.  of  wire  0.05  of  an  ohm.      This  is  at  the 
rate  of  0.25  of  an  ohm  per  1,000  ft.     The  difference  between 
this  desired  resistance  and  that  of  200  ft.  No.  4  of  copper  wire 
is  inconsiderable  and  may  be  ignored.     In  practice,  a  differ- 
ence like  that  of  Exercise  6  would  probably  be  ignored,  and 
No.  24  wire  used. 

8.  E  =  CR  =  10  x  3.8  =  38. 

9.  The  total  resistance  of  the  lamp,  including  the  arc,  is 
3.8  ohms  +  0.62  of  an  ohm  =  4.42  ohms. 

E  =  CR  =  10  x  4.42  =  44.2. 

10.  206  -r-  (1.6  +  25.4)  =  7.63,  the  number  of  amperes. 

11.  (2.8  +  1.1  -f  9.36)  x  14.8  =  196.248. 

12.  It  is  not  running  .fast  enough.     With  the  given  resist- 
ances, a  25-ampere  current  will  require  an  E.M.F.  of  212.5 
volts.     The  dynamo  must  be  "  speeded  up  "  so  as  to  give  the 
additional  12.5  volts. 


94  KEY  TO   THE   PROBLEMS 

13.  81.58  -r-  29.67  =  2.75,  the  total  number  of  ohms. 
2.75  ohms  —  1.14  ohms  =  1.61  ohms. 

14.  The  total  resistance  of  the  circuit  is   (157.5  -*•  17.5  =) 
9  ohms.     9  ohms  —  4.58  ohms  =  4.42  ohms. 

15.  (39.3  x  3  +  11.2)  x  1.2  =  154.92. 

16.  As  the  lamps  were  in  series,  the  strength  and  E.M.F.  of 
the  current  must  have  been  the  same  in  both  lamps. 

97  x  2  +  12  =  206,  the  number  of  ohms. 
E  =  CR  =  1  X  206  =  206,  the  number  of  volts. 

17.  The  armature  wire  must  carry  4.8  amperes.     4.8  -f-  2,000 
=  0.0024,  the   required   area  in  square  inches.     The  nearest 
size,  as  given  in  the  fifth  column  of  the  table  on  page  593,  is 
No.  15. 

18.  The  E.M.F.  required  to  overcome  the  re- 
sistance of  the  lamps  is  45  volts  x  60  =  2,700     volts. 

The  E.M.F.  required  to  overcome  the  other 
resistance  is  9.6  (16  +  5)  =  201.6  volts. 

Total  voltage  required,     2,901.6  volts. 

7,200,000  x  120  x  780  =  112Q2 
60  x  100,000,000 

20.  It  would  lessen  it. 

21.  The  a,rmature  that  has  twice  as  many  bobbins  will  develop 
twice  as  great  an  E.M.F. 

22.  As  the  external  resistance  is  lessened,  the  current  will 
be  increased ;  if  such  increase  is  continued  too  far,  the  current 
will  become  greater  than  the  No.  8  wire  can  safely  carry,  the 
wire  will  become  red-hot,  and  its  insulation  destroyed ;  i.e.,  the 
armature  will  "  burn  out."     Hence  the  danger  of  short-circuit- 
ing a  dynamo. 

23.  An  increase  of  the  resistance  in  the  field  circuit  of  a 
shunt  dynamo  lessens  the  current  that  passes  around  the  field 


IN  AVERY'S  SCHOOL  PHYSICS.  95 

magnets,  weakens  the  magnetic  field,  and  lessens  the  number 
of  lines  of  force  that  pass  through  the  armature.  The  E.M.F. 
of  such  dynamos  is  often  thus  controlled. 

24.  To  send  a  2-ampere  current  through  a  2-ohm  coil  requires 
a  pressure  of  4  volts  ;  E  =  CR.  The  dynamo  must  be  run  at 
a  speed  that  will  give  this  voltage.  The  added  resistance  must 
be  such  that  (9.6  —  2=)  7.  6  amperes  will  flow  through  it  at 
this  pressure.  ^  . 


or  7.6:  2::  2:  a;;    x  =  0.526. 

25.   (a)  110-5-0.96  =  114.58  +  . 

(6)    The  fall  of  potential  in  the  line  =  114.58  volts  -  110 
volts  =  4.58  volts. 

Current  required  =  127  x  0.5  =  63.5,  the  number  of  amperes. 

=  0.0721  +,  the  number  of  ohms. 


63.  5 

(c)  0.0721  -f-  1,240  =  0.0000581,  the  number  of  ohms. 

(d)  No.  000,  i.e.,  "three  naught"  wire. 

C 

26.  R  =  —  .     The  resistance  of  the  wire  must  be  (110  -5-  15  =) 

7.333  ohms.  The  table  shows  that  the  resistance  of  a  foot  of 
No.  14  wire  is  0.002585  of  an  ohm.  7.333  -s-  0.002585  =  2,837, 
the  number  of  feet. 

27.  (a)  The  two  miles  of  wire  have  a  resistance  of 

(0.051  x  5.28  x  2  =)  0.53856  of  an  ohm. 

(b)  175  -j-  350  =  0.5,  the  number  of  amperes  required  for 
each  lamp. 

0.5  x  100  =  50,  the  number  of  amperes  required  for  the  100 
lamps. 

E  =  CR  =  50  x  0.53856  =  26.928,  the  fall  of  potential  or 
loss  of  voltage  due  to  the  resistance  of  the  wire. 

110  +  26.928  =  136.928,  the  voltage  required  at  the  dynamo. 


96          KEY   TO   PROBLEMS  IN   AVERY'S   PHYSICS. 

28.  As  either  sounder  adds  little  relatively  to  the  high  resist- 
ance of  the  line,  the  current  strength  is  nearly  as  great  with 
the  high  resistance  sounder  in  the  circuit  as  it  is  with  the 
other.     On  the  other  hand,  the  difference  in  the  number  of  the 
turns   of  the   wire   does    materially   affect   the   ampere-turns 
(§  376)  in  favor  of  the  high  resistance  instrument.     Do  not  let 
the  pupil  get  the  idea  that  the  high  resistance  of  the  instru- 
ment is  the  cause  of  the  greater  efficiency.     The  resistance  is 
an  incidental  and  not  a  desirable  factor  in  the  case,  but  in 
order  to  get  the  great  number  of  turns  in  a  small  space,  a  long 
fine  wire  is  necessary,  and  this  involves  a  high  resistance. 

29.  See  §  424. 

30.  The  difficulty  in  the  latter  case  is  due  to  the  alternating 
nature  of  the  telephonic  current,  and  to  the  greater  resistance 
and  impedance  of  the  circuit. 


° 


YB   16958 


926494 


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